Answer:
The force needed to launch the projectile is 150000 N.
Explanation:
We can find the force using the following equation:
[tex] F = ma [/tex]
Where:
m: is the mass = 15 kg
a: is the acceleration
First, we need to find the acceleration of the projectile:
[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]
Where:
[tex]v_{f}[/tex]: is the final speed
[tex]v_{0}[/tex]: is the initial speed = 0
x: is the distance = 2 m
The final speed is:
[tex]v_{f} = \frac{1 km}{5 s}*\frac{1000 m}{1 km} = 200 m/s[/tex]
Then, the acceleration is:
[tex]a = \frac{v_{f}^{2}}{2x} = \frac{(200 m/s)^{2}}{2*2 m} = 10000 m/s^{2}[/tex]
Finally, the force is:
[tex]F = ma = 15 kg*10000 m/s^{2} = 150000 N[/tex]
Therefore, the force needed to launch the projectile is 150000 N.
I hope it helps you!
A man exerts a force on a large couch by pushing on it. Which of the following must be a true statement about the couch after he begins pushing?
A. The couch will maintain a constant acceleration due to the man's push.
B. The couch will remain at rest.
C. The couch will remain at rest if the push does not overcome the force of friction.
D. The couch will accelerate and then remain at a constant velocity.
Answer: the answer is c
Explanation:
What is Newtons third law of motion?
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction. Explanation:
Answer:
His third law states that for every action (force) in nature there is an equal and opposite reaction
Explanation:
In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. Notice that the forces are exerted on different objects.
A 1800 kg car moves along a horizontal road at speed v₀ = 18.4 m/s. The road is wet, so the static friction coefficient between the tires and the road is only µ static = 0.188 and the kinetic friction coefficient is even lower, µ kinetic = 0.1316. The acceleration of gravity is 9.8 m/s². What is the shortest possible stopping distance for the car under such conditions? (neglect the reaction time of the driver and round your answer to 4 decimal places)
Answer:
The shortest possible distance is [tex]|s| = 91.9 \ m[/tex]
Explanation:
From the question we are told that
The mass of the car is [tex]m = 1800 \ kg[/tex]
The speed along the horizontal road is [tex]v_o =u = 18.4 \ m/s[/tex]
The static friction coefficient is [tex]\mu_s = 0.188[/tex]
The kinetic friction coefficient is [tex]\mu_k = 0.1316[/tex]
Generally the static frictional force acting on the car is mathematically represented as
[tex]F_f = m * g * \mu_s[/tex]
Generally the force propelling the car is mathematically represented as
[tex]F = m * a[/tex]
Here a is the maximum acceleration
at the point which the car stops ,
[tex]F = F_f[/tex]
=> [tex]m * g * \mu_s = ma[/tex]
=> [tex]g * \mu_s =a[/tex]
=> [tex]a = 9.8 * 0.188[/tex]
=> [tex]a = 1.8424 \ m/s^2[/tex]
Generally from kinematic equation
[tex]v^2 = u^2 + 2as[/tex]
Here v is the final velocity of the car and the value is zero given that the car comes to rest
So
[tex]0^2 = 18.4^2 + 2* 1.8424 s[/tex]
=> [tex]s = - \frac{18.4^2}{2 * 1.8424}[/tex]
=> [tex]|s| = 91.9 \ m[/tex]
pls answer quick & an accurate answer will get brainliest
Answer:
3.766 kg or 3.7 kg
Explanation:
F= M*A equation
True or False: A cloud’s only purpose is to create precipitation.
Answer:
I think it's false
Explanation:
clouds also help regulate the Earth's energy balance by reflecting and scattering solar radiation and by absorbing the Earth's infrared radiation.
Answer:
false
Explanation:
Which term best explains why you hear an echo?
Answer:
reflection
Explanation:
Answer:
reflection
Explanation:
A p e x
A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235m below. If the plane is traveling horizontally with a speed of 250 km/h (69.4m/s), (a) how far in advance of the recipients ( horizontal distance) must the goods be dropped. (b) Suppose, instead that the plane release the supplies a horizontal distance of 425m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climber's position? (c) with what speed do the supplies land in the latter case
Answer:
Explanation:
For vertical displacement of food-supply
if time taken to reach the ground be t
235 = 1/2 g t² , initial velocity downwards is zero.
t = 6.92 s
Horizontal displacement of food during this period
= 6.92 x 69.4 m
= 480.25 m .
b )
time required to cover horizontal distance of 425 m
= 425 / 69.4 = 6.124 s
This time period will be the time of vertical fall of 235 m . Let initial vertical velocity required be u
h = ut + 1/2 gt²
235 = u x 6.124 + .5 x 9.8 x 6.124²
235 = u x 6.124 + 183.76
u = 8.36 m /s
c )
v = u + gt
= 8.36 + 9.8 x 6.124
= 68.37 m /s
This will be vertical component of velocity .
horizontal velocity = 69.4 m /s
resultant velocity = √ ( 68.37² + 69.4²)
= √(4674.45 +4816.36)
= 97.42 m /s
Conduction In Everyday Life Examples
at least 5 plz help
1. Heat transfer from a hot burner on the stove into a pot or pan
2. A metal spoon that gets hot from the boiling water
3. Chocolate candy in your hand that melts because of the heat
4. The wires in your house that conduct electricity
5. Roasting marshmallows, the heat will soften the marshmallow
Answer:
the engine heat from a car you just turned on.
a heater
a pad warmer
a hot pot
using a campfire to roast hot dogs
How much air resistance acts on a 100-N bag of nails that falls at its terminal speed?
A car’s velocity changes from 35 m/s to stopped in 13 seconds. Calculate
acceleration.
Answer:
Acceleration = 3m/s^2
Vf= 0 Vi =35m/s t= 13s
Explanation:
[tex]Acceleration = \frac{Change in velocity}{Change in time}\\ = \frac{35m/s}{13s}\\ a = 2.69m/s^2\\ a = 2.7m/s^2\\ a = 3m/s^2[/tex]
Una larga fanja de pavimento tiene marcas a intervalos de 10m. Los estudiantes usan cronometros para registrar los tiempos en que un automovil pasa por cada marca. Asi han obtenido los datos siguientes: Distancia,m l 0 l 10 l 20 l 30 l 40 l 50 l Tiempo,s l 0 l 2.1 l 4.2 l 6.3 l 8.4 l 10.5 l A) Cual es la rapidez media del vehiculo? B) Al cabo de cuanto tiempo la distancia es igual a 34m? C) Cual es la aceleracion del automovil?
Answer:
men
Explanation:
The chart shows the percentage of different elements in the human body.
Which element is the most prevalent in the human body?
nitrogen
hydrogen
carbon
oxygen
A 1500 kg car is parked at the top of a hill 5.2 m high. What is the velocity of the car, in meters per second, when it reaches the bottom of the hill?
Answer:
Explanation:
The car will fall with acceleration due to gravity which is equal to 9.8 m /s²
For downward fall ,
initial velocity u = o
acceleration due to gravity = g = 9.8 m /s
final velocity v = ?
displacement h = 5.2 m
v² = u² + 2 gh
v² = 0 + 2 x 9.8 x 5.2
= 101.92
v = 10.1 m /s
What fonts do you use to create color coded lyrics
HELPP ITS DUE IN 5 MINUTES FREE BODY DIAGRAMS
Answer:
I think it's part c . but sorry if its wrong
Answer the following questions
Answer:
9 - 10N to the left
10 - There is no change on the object
Explanation:
Can I have brainliest answer pls?
Question 7 (2 points)
Rachel performed an experiment testing the hours students slept with their
performance on a test. In this experiment, the hours that they slept was the_____ variable, while the grade they got on
the test was the
_____ variable.
A model rocket Was built last weekend. We launched it in a nearby field. The rocket took off Straight up with an initial velocity 62.3 m/s. If it accelerates upward at 4.30 m/s2 until the engines stop 3.73 s later: (a) What is the maximum height the rocket reaches? (b) How long is the rocket in the air? (c) What is the velocity of the rocket as it hits the ground?
Answer:
a) y_total = 575.4 m , b) t_total = 22.56 s , c) v = 106.20 m
Explanation:
a) We can solve this exercise using the scientific relations
Let's find the height reached while the engines are running
v₁ = v₀ + a t
v₁ = 62.3 + 4.30 3.73
v₁ = 78.339 m / s
in this time it rose
y₁ = v₀ t₁ + ½ a t₁²
y₁ = 62.3 3.73 + ½ 4.30 3.73²
y₁ = 262.29 m
At this moment the engines are turned off and the rocket continues in a vertical launch, suppose that the acceleration of gravity is constant in this path
v₂ = v₁ 2 - 2 g y₂
at maximum height the velocity is zero (v₂ = 0)
y₂ = v₁ 2 / 2g
y₂ = 78.339 2/2 9.8
y₂ = 313.11 m
the total height is
y_total = y₁ + y₂
y_total = 262.29 + 313.11
y_total = 575.4 m
b) the time the rocket is in the air is
t_total = t₁ + t₂
where t₂ is the time after the engines have shut down
y = y₁ + v₁ t₂ - ½ g t₂²
y = 0
0 = 262.29 + 78.339 t₂ - ½ 9.8 t₂²
t₂² - 15.99 t₂ - 53.53 = 0
t₂ = [15.99 ±√ (15.99² +4 53.53)] / 2
t₂ = [15.99 ± 21.67] / 2
t₂´ = 18.83 s
t₂´´ = -2.84 s
since the time must be a positive quantity the correct result is t2 = 18.83 s, the total time in the air is
t_total = 3.73 + 18.83
t_total = 22.56 s
c) the speed when it hits the ground
We can perform this calculation starting with the maximum height y = 575.4 m where it has zero initial velocity (vo = 0)
v² = v₀² + 2 g y
v = √ 2gy
v = √ (2 9.8 575.4)
v = 106.20 m
Where does digestion begin?
Answer:
Digestion begins in the mouth, when you chew.
Which evidence did Alfred Wegener’s original theory of continental drift have access to?
Answer:
Evidence for continental drift
Wegener knew that fossil plants and animals such as mesosaurs, a freshwater reptile found only South America and Africa during the Permian period, could be found on many continents. He also matched up rocks on either side of the Atlantic Ocean like puzzle pieces.
Explanation:
gold has a density of 19300 kg/m3 calculate the mass of 0.02m3 of gold in kilograms
Mass if a substance is the product of its volume and density. The mass of gold of 0.02 m³ with a density of 19300 kg/m³ is 386 kg.
What is density?Density of a substance is the measure of its mass per unit volume. Thus, it says how much denser the object is in a given volume. Density of a substance is dependent on its bond type, temperature and pressure beside the volume and mass.
Volume can be defined as the space occupied by the substance. Larger the volume , less dense the substance is. However as the mass increases volume also increases.
Mass of an object is the product of its volume and density.
Given the volume of gold = 0.02 m³
density = 19300 kg/m³.
mass = volume × density
= 0.02 m³ × 19300 kg/m³
= 386 kg
Therefore, the mass of gold is 386 Kg.
To find more on gold, refer here:
https://brainly.com/question/11405288
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A +4.0 uC charge is placed on the x axis at x= +3.0 m, and a -2.0 uC is located on the y-axis at y= -1.0 m. Point A is on the y axis at y= +4.0 m. Determine the electric potential at point A (relative to zero at the origin).
Answer:
The potential is [tex]V_A = 9600 \ V[/tex]
Explanation:
From the question we are told that
The magnitude of the charge is [tex]q_1 = 4 \mu C = 4*10^{-6} \ C[/tex]
The position of the charge is [tex]x = + 3.0 \ m[/tex]
The magnitude of the second charge is [tex]q_2 = -2.0 \mu C = -2.0 *10^{-6} \ C[/tex]
The position is [tex]y_1 = - 1.0 \ m[/tex]
The position of point A is [tex]y_2 = + 4.0 \ m[/tex]
Generally the electric potential at A due to the first charge is mathematically represented as
[tex]V_a = \frac{k * q_1 }{r_1 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_1[/tex] is the distance between first charge and a which is mathematically represented as
[tex]r_1 = \sqrt{x^2 + y_2 ^2 }[/tex]
=> [tex]r_1 = \sqrt{3^2 + 4 ^2 }[/tex]
=> [tex]r_1 = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * 4*10^{-6} }{5 }[/tex]
[tex]V_a = 7200 \ V[/tex]
Generally the electric potential at A due to the second charge is mathematically represented as
[tex]V_b = \frac{k * q_2 }{r_2 }[/tex]
Here k is the coulombs constant with value [tex]k = 9*10^{9} \ \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}[/tex]
[tex]r_2[/tex] is the distance between second charge and a which is mathematically represented as
[tex]r_2 = y_2 - y[/tex]
=> [tex]r _2 = 4.0 - (-1.0)[/tex]
=> [tex]r = 5 \ m[/tex]
So
[tex]V_a = \frac{9*10^9 * -2*10^{-6} }{5 }[/tex]
[tex]V_a = -3600 \ V[/tex]
So the net potential difference at point A due to the charges is mathematically represented as
[tex]V_n = V_a + V_b[/tex]
=> [tex]V_n = 7200 - 3600[/tex]
=> [tex]V_n = 3600 V[/tex]
Generally the net potential difference at the origin due to both charges is mathematically represented as
[tex]V_N = V_c + V_d[/tex]
Here
[tex]V_c = \frac{k * q_1 }{x}[/tex]
=> [tex]V_c = \frac{9*10^9 * 4*10^{-6} }{3}[/tex]
=> [tex]V_c = 12000 V[/tex]
and
[tex]V_d= \frac{k * q_2 }{y}[/tex]
=> [tex]V_c = \frac{9*10^9 * -2*10^{-6} }{1}[/tex]
=> [tex]V_c =- 18000 V[/tex]
Generally the net potential difference at the origin is
[tex]V_N = 12000 - 18000[/tex]
=> [tex]V_N = -6000[/tex]
Generally the potential difference at A relative to zero at the origin is mathematically evaluated as
[tex]V_A = V_n - V_N[/tex]
=> [tex]V_A = 3600 - (-6000)[/tex]
=> [tex]V_A = 9600 \ V[/tex]