Answer:
5 cm - 53 degrees
Explanation:
[tex] \sqrt{ {4}^{2} + {3}^{2} } = 5cm[/tex]
[tex] \tan ^{ - 1} (4 \div 3) [/tex] = 53°
What can you infer from the fact that metals are good conductors of electricity?
Answer:
Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.
A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?
Answer:
The maximum angular velocity is 20 rad/s
Explanation:
Given;
torque, τ = 10 N
maximum mechanical power, P = 200 J/s
The output power of the pmdc is given as;
P = τω
where;
P is the maximum mechanical power
ω is the maximum angular velocity
ω = P / τ
ω = (200) / (10)
ω = 20 rad/s
Therefore, the maximum angular velocity is 20 rad/s
During a hockey game, a puck is given an initial speed of 10 m/s. It slides 50 m on the horizontal ice before it stops due to friction. What is the coefficient of kinetic friction between the pick and the ice.A) 0.12B) 0.10C) 0.11D) 0.090
Answer:
The value is [tex]\mu_k = 0.102[/tex]
Explanation:
From the question we are told that
The initial speed of the pluck is [tex]u = 10 \ m/s[/tex]
The distance it slides on the horizontal ice is [tex]s = 50 \ m[/tex]
Generally from kinematic equation we have that
[tex]v^2 = u^2 + 2as[/tex]
Here v is is the final velocity and the value is 0 m/s given that the pluck came to rest, so
[tex]0^2 = 10 ^2 + 2* a * 50[/tex]
=> [tex]a = - 1 \ m/s^2[/tex]
Here the negative sign show that the pluck is decelerating
Generally the force applied on the pluck is equal to the frictional force experienced by the pluck
So
[tex]F = F_f[/tex]
=> [tex]m * a = m* g * \mu_k[/tex]
=> [tex]1 = 9.8 * \mu_k[/tex]
=> [tex]\mu_k = 0.102[/tex]
An aluminum baking sheet with a mass of 225 g absorbs 2.4 x 104 J from an oven. If its temperature was initially 25 C, what will its new temperature be?
Answer:
The value is [tex]T_2 =416.9 \ K[/tex]
Explanation:
From the question we are told that
The mass of the aluminum baking sheet is [tex]m = 225 \ g = 0.225 \ kg[/tex]
The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]
The initial temperature is [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]
Generally the heat absorbed is mathematically represented as
[tex]Q = m * c_a * [T_2 - T_1][/tex]
Here [tex]c_a[/tex] is the specific heat capacity of aluminum with value [tex]c_a = 897 \ J / kg \cdot K[/tex]
So
[tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]
=> [tex]T_2 - 298 = 118.915[/tex]
=> [tex]T_2 =416.9 \ K[/tex]
which factor does not affect the strength of an electromagnet
Answer:
the placement of the ammeter in the circuit
Explanation:
A 2.80 kg mass is dropped from a height of 4.50 m. Find its potential energy when it reaches the ground.
Answer:
123.48J
Explanation:
Given parameters:
Mass of the ball = 2.8kg
Height = 4.5m
Unknown:
Potential energy = ?
Solution:
The potential energy is the energy due to the position of a body. It is mathematically given as;
P.E = mgh
m is the mass
g is the acceleration due to gravity
h is the height
Now insert the parameters and solve;
P.E = 2.8 x 4.5 x 9.8 = 123.48J
Answer:
0
Explanation:
There is 0 PE when its on the ground
A particular engine has a power output of 2 kW and an efficiency of 27%. If the engine expels 9085 J of thermal energy in each cycle, find the heat absorbed in each cycle. Answer in units of J.
Answer:
12445 J
Explanation:
Given that
Power output, P = 5 kW
efficiency of the engine, e = 27% = 0.27
Thermal energy expelled, Q(c) = 9085 J
Heat absorbed, Q(h) = ?
Using the formula
e = W/Q(h)
e = [Q(h) - Q(c)] / Q(h)
e = 1 - Q(c)/Q(h)
Now, substituting the values into the formula, we have
0.27 = 1 - 9085/Q(h)
9085/Q(h) = 1 - 0.27
9085/Q(h) = 0.73
Q(h) = 9085 / 0.73
Q(h) = 12445 J
Thus, the heat absorbed is 12445 J
ANSWER THIS FOR 16 POINTS!!!!!!!!!!
When hitting the ski slopes when does the skier has the most potential energy??
Answer:
As the ski jumper starts moving downhill, some of his potential energy changes into kinetic energy (KE). Kinetic energy moves him down the slope to the ramp. When the ski jumper takes off from the ramp, some of his kinetic energy is changed back into potential energy as he rises in the air.
Explanation: hope this helps
Answer:
at the top of the slope
Explanation:
What differentiates galaxy groups from clusters?
A.
Clusters are bigger than groups.
B.
Clusters are more massive than groups.
C.
Clusters contain a hot intracluster medium, whereas groups do not.
D.
Clusters are collections of galaxy groups, whereas groups are collections of galaxies.
E.
Clusters don't gravitationally bind galaxies together, while groups bind galaxies gravitationally.
Answer:
A
Explanation:
Galaxy clusters are basically very large (>50 galaxies) groups
Answer:
The correct answer would be:
C.
Clusters contain a hot intracluster medium, whereas groups do not.
#PLATOFAM
Have a nice day!
How much of the day do we spend sleeping?
Answer:
roughly 8-10 hours for the average human being
Explanation:
A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?
Answer:
bus momentum
p_bus= m_bus x v_bus
=18,200 x 16.5
basball momentum
pball=mball x vball
=0.142 x v
p_bus = pball
18200 x 16.5 = 0.142 x v
v=(18200 x 16.5)/0.142
v is the answer for baseball
Explanation:
⚠️not my answer tryna be honest here⚠️
The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶ m/s.
What is momentum?Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.
Given that one bus is having a mass of 18200 Kg and 13.2 m/s speed. The momentum is:
p = mv
=18200 kg × 13.5 m/s
= 245700 Kg m/s
To have a momentum of 245700 Kg m/s the base ball with 0. 142 g have to have a velocity = 245700 Kg m/s / 0.142 g
=1.7 × 10 ⁶ m/s
Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶ m/s
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Two cars each have a mass of 1050 kg. If the gravitational force between
them is 2.27 x 10-7N, how far apart are they? G = 6.67 10-11 N:(m/kg)2
A. 21 m
B. 5.6 m
C. 33 m
D. 18 m
DUBMIT
Answer: D.
Explanation:
d = ± √ G m 1 m 2 /F
Because distance cannot be negative there it goes:
d = √ G m 1 m 2 /F
d = √ 6.67 ⋅ 10 − 11 ⋅ 1050 ⋅ 1050 2.27 ⋅ 10 − 7
d = √ 0.0000735 2.27 ⋅ 10 − 7
d = √ 3239504.405
d ≈ 1799.86 m = 18m
A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide
Answer:
[tex]\theta = 16.70 ^{\circ}[/tex]
Explanation:
The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.
[tex]\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta[/tex]Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.
Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.
[tex]w=mg[/tex]We are given the mass of the block (kg) and we know that g = 9.8 m/s².
[tex]w=(5)(9.8) = 49 \ \text{N}[/tex]Now we can use this force in the equation:
[tex]\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}[/tex]Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.
[tex]\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}[/tex] [tex]0.30=\text{tan} \theta[/tex]Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].
Evaluate this equation by taking the inverse tangent of both sides of the equation.
[tex]\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)[/tex] [tex]\text{tan}^-^1 (0.30) =\theta[/tex] [tex]\theta = 16.69924423[/tex]The minimum angle at which the block will begin to slide is about 16.70 degrees.
to determine the height of a steep cliff an experimenter stations a sensor on the top of the cliff then fires a pellet vertically upward with an initial velocity of 80 m/s . the sensor reports that the pellet reached a maximum height 3 meters above the edge of the cliff. how high is the cliff?
a. 77 m
b. 237 m
c. 317 m
d. 637 m
e. 797 m
Answer:
c. 317 m
Explanation:
Vertical Launch Upwards
It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.
If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:
[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]
We'll assume the acceleration of gravity as [tex]g=10\ m/s^2[/tex].The pellet is vertically upward launched with vo=80 m/s. The maximum height is:
[tex]\displaystyle h_m=\frac{80^2}{2*10}=320[/tex]
[tex]h_m = 320\ m[/tex]
That height is 3 meters above the edge of the cliff, thus the cliff is 320-3=317 m hight
c. 317 m
Part A:
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
Part B:
A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part C:
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part D:
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?
Part E:
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
Part F:
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?
Part G:
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.
Answer:
a) 0 V
b) 10 turns
c) 4000 turns
d) 12.5 A
e) 400 W
f) 0.5 A
g) 95.4%
Explanation:
A
0
B
To solve this, we would be using the simple relationship between voltage and number of turns
V1/V2 = N1/N2
500/25 = 200/N2
20 = 200/N2
N2 = 200/20
N2 = 10 turns
C
Here also, we would be using the relationship between current and the number of turns
I1/I2 = N2/N1
500/25 = N2/20
20 = N2/20
N2 = 20 * 20
N2 = 4000 turns
D
Like in the previous question, current and the number of turn relationship is used
N1/N2 = I2/I1
400/80 = I2/2.5
5 = I2/2.5
I2 = 5 * 2.5
I2 = 12.5 A
E
The power remains unchanged at 400 W
F
Power = Voltage * Current
P = VI
I = P/V
I = 60/120
I = 0.5 A
G
95.4%
The transformer is a device used to step up or step down voltage.
Part A;
Given that;
Es/Ep = Ns/Np
Es = voltage in the secondary coil
Ep = voltage in primary coil
Ns = Number of turns in secondary coil
Np = Number of coils in primary coil
Es = Ns/Np × Ep
Es = 200/100 × 1.5 V
Es = 3 V
Part B
Ns = Es/Ep × Np
Ns = 25/500 × 200
Ns = 10 turns
Part C
Ns/Np = Ip/Is
Ns = Ip/Is × Np
Ns = 500/25 × 200
Ns = 4000 turns
Part D
Ns/Np = Ip/Is
NsIs = NpIp
Is = NpIp/Ns
Is = 400 × 2.5/80
Is =12.5 A
Part E
The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.
Part F
Power supplied = 60 watts
Voltage of primary coil = 120 V
Since;
P = IV
I = P/V = 60/120 = 0.5 A
Part G
Since;
E = 100Pout/Pin
Pin = 120 V × 2 A = 240 W
Pout = 19.4 V × 11.8 A = 228.92 W
E = 100(228.92/240)
E = 95.4%
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Bob Beamon's 1968 Olympic long jump set a world record which remains unbroken to this day. This amazing jump resulted from an initial velocity of 9.5 m/s at an angle of 40 degrees from the horizontal.
1. Calculate the initial horizontal velocity (V_ix) to two significant figures:
2. Calculate the initial vertical velocity (V_iy) to two significant figures:
3. Calculate the time needed to reach the highest point of the jump (t_1/2) to two significant figures:
4. Calculate the total time (t_TOT) needed to complete the jump to two significant figures:
5. Calculate the maximum height (h) reached during the jump to two significant figures:
6. Calculate the range (total horizontal distance) of his jump to two significant figures:
Please answer today! Thanks!
Answer:
1.) 7.3 m/s 2.) 6.1 m/s
Explanation:
To calculate the initial horizontal velocity with just degrees and velocity alone is pretty simple. The formula is Velocity*cos(degrees)
eg 9.5*cos(40)
2. To calculate the initial vertical velocity with just degrees and velocity alone is pretty simple. The formula Velocity*sin(degrees)
eg 9.5*sin(40)
A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?
Answer:
600 KPa.
Explanation:
From the question given above, the following data were obtained:
Initial volume (V1) = 0.075 m³
Final volume (V2) = 0.45 m³
Final pressure (P2) = 100 KPa
Initial pressure (P1) =?
Temperature = constant
The initial pressure can be obtained by using the Boyle's law equation as shown below:
P1V1 = P2V2
P1 × 0.075 = 100 × 0.45
P1 × 0.075 = 45
Divide both side by 0.075
P1 = 45 / 0.075
P1 = 600 KPa.
Thus, the initial pressure in the balloon is 600 KPa.
A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]
[tex]\\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]
[tex]\\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]
[tex]\\[/tex]
☯ For left penetration (s₂)
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]
[tex]\\[/tex]
[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]
A stone dropped from a bridge strikes the water 5.6 seconds later. What is the final velocity in meters/s?
A) 179.78 meters/s
B) 5.71 meters/s
C) 1.75 meters/s
D) 54.88 meters/s
Answer: 54.88 meters/s
Explanation:
The final velocity will be calculated by using the formula:
v = u + at
where,
v = final velocity
u = initial velocity = 0
a = 9.8
t = 5.6
Therefore, we slot the value back into the formula. This will be:
v = u + at
v = 0 + (9.8 × 5.6)
v = 0 + 54.88
v = 54.88 meters per second
Therefore, the final velocity is 54.88m/s
10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)
Answer: mass = 48.47 kg.
Explanation:
Formula : Weight = mg , where m = mass of body , g= acceleration due to gravity .
Given: Weight = 475 N
[tex]g= 9.8\ m/s^2[/tex]
Substitute all values in formula , we get
[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]
Hence, his mass = 48.47 kg.
The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times
Answer: The correct answer is C
Explanation:
A motorcycle skids for a distance of 2.0 m with the icy road pushing on its tires with force of 120 N as its
brakes are applied
What is the change in kinetic energy for the motorcycle?
Round the answer to two significant digits.
Answer:
-240
Explanation:
A motorcycle skids for a distance of 2.0 m on an icy road, then the change in kinetic energy for the motorcycle will be equal to -240 J.
What is kinetic energy?The force which a moving object has is referred to as kinetic energy in physics. It is defined as the number of effort required to propel a person of a specific mass from still to a specific velocity.
Aside from slight fluctuations in speed, your body holds onto the kinetic energy it obtains during acceleration.
When the body slows down from its present level to a condition of rest, the same quantity of energy is used.
Formally, kinetic energy is any quantity that has a gradient concerning time in the Lagrangian of a system.
As per the given information in the question,
Distance, d = 2.0 m
Friction, f = 120 N
The angle between displacement and friction force, θ = 180°
Now, the change in kinetic energy for the motorcycle = Work done by the friction.
K.E = f × d(cos θ)
= 120 (2.0 m)(cos 180°)
Δ K.E = -240 J
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What causes hot air to rise from the pavement on a hot summer day?
Answer:
radation because the heat gets too intense and is forced to rise
The image below shows four boxes that each contain a different sample of gas. The atoms of each gas are represented by dots, 1 2 3 4 Which box contains the gas with the greatest density?
A. 1
B. 2
C. 3
D. 4
A 950 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72 m, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?
Compute the car's weight:
W = m g = (950 kg) (9.8 m/s²) = 9310 N
The net vertical force on the car is
∑ F = N - W = 0
so the normal force has magnitude
N = W = 9310 N
Then the friction force that keeps the car from skidding has magnitude F = µ N, where µ is the coefficient of friction, and it's friction that makes up the net horizontal force on the car. By Newton's second law, we have
F = m a
µ N = m v ² / R
µ (9310 N) = (950 kg) (25 m/s)² / (72 m)
µ ≈ 0.89
A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?
[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]
[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]
[tex]\\[/tex]
[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]
☯ As we know that,
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]
[tex]\\[/tex]
[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]
why do feet smell and noses run?
Answer:
Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.
Explanation:
Hope this helps
A speeding race car primarily contains potential energy.
:True
False
g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion
Answer:
The answer is "1557 meters".
Explanation:
speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]
[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]
A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?
Answer:
Δx = 39.1 m
Explanation:
Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:[tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]
where vf is the final velocity (0 in our case), v₀ is the initial velocity
(25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance
traveled since the brakes are applied.
Solving (1) for Δx, we have:[tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]
The car will travel a distance of 39.1 m before its stops.
To solve the problem above, use the equations of motion below.
Equation:
v² = u²+2as................... Equation 1Where:
v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance coveredFrom the question,
Given:
v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)Substitute these values into equation 1
⇒ 0² = 25²+2(-8)(s)Solve for s
⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 mHence, The car will travel a distance of 39.1 m before its stops.
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