Answer:
The velocity at which the ball leave the ground is 6.62 m/s.
Explanation:
Given;
mass of tennis ball, m = 60 g = 0.06 kg
height at which the ball fell, h = 2.9 m
percentage of kinetic energy lost, = 23%
The potential energy of the tennis ball = mgh = 0.06 x 9.8 x 2.9
= 1.7052 J
As the ball hits the ground, the potential energy of the ball is converted to kinetic energy.
K.E = 1.7052 J
Kinetic energy lost = 0.23 x 1.7052
= 0.392 J
The remaining kinetic energy ΔK.E = 1.7052 J - 0.392 J
= 1.3132 J
The velocity at which the ball leave the ground is calculated as;
ΔK.E = ¹/₂mv²
[tex]v = \sqrt{\frac{2\Delta K.E}{m} } \\\\v = \sqrt{\frac{2\times 1.3132}{0.06} }\\\\v = 6.62 \ m/s[/tex]
Therefore, the velocity at which the ball leave the ground is 6.62 m/s.
A vertical straight wire 35.0 cmcm in length carries a current. You do not know either the magnitude of the current or whether the current is moving upward or downward. If there is a uniform horizontal magnetic field of 0.0300 TT that points due north, the wire experiences a horizontal magnetic force to the west of 0.0180 NN. Find the magnitude of the current.
Answer:
[tex]1.714\ \text{A}[/tex]
Explanation:
F = Magnetic force = 0.018 N
B = Magnetic field = 0.03 T
L = Length of wire = 35 cm
[tex]\theta[/tex] = Angle between current and magnetic field = [tex]90^{\circ}[/tex]
Magnetic force is given by
[tex]F=IBL\sin\theta\\\Rightarrow I=\dfrac{F}{BL\sin\theta}\\\Rightarrow I=\dfrac{0.018}{0.03\times 35\times 10^{-2}\times \sin90^{\circ}}\\\Rightarrow I=1.714\ \text{A}[/tex]
The magnitude of the current is [tex]1.714\ \text{A}[/tex].
To understand the meaning and possible applications of the work-energy theorem. In this problem, you will use your prior knowledge to derive one of the most important relationships in mechanics: the work-energy theorem. We will start with a special case: a particle of mass m moving in the x direction at constant acceleration a. During a certain interval of time, the particle accelerates from vi to vf, undergoing displacement s given by s=xf−xi.
A. Find the acceleration a of the particle.
B. Evaluate the integral W = integarvf,vi mudu.
Answer:
a) the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Explanation:
Given the data in the question;
force on particle F = ma
displacement s = x[tex]_f[/tex] - x[tex]_i[/tex]
work done on the particle W = Fs = mas
we know that; change in energy = work done { work energy theorem }
[tex]\frac{1}{2}[/tex]m( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = mas
[tex]\frac{1}{2}[/tex]( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = as
( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) = 2as
a = ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
Therefore, the acceleration of the particle is ( v[tex]_f[/tex]² - v[tex]_i[/tex]² ) / 2as
b) Evaluate the integral W = [tex]\int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W = \int\limits^{v_{f} }_{v_{i} } mvdv[/tex]
[tex]W =m[\frac{v^{2} }{2} ]^{vf}_{vi}[/tex]
W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
Therefore, the integral W = [tex]\frac{1}{2}[/tex]m( [tex]_f[/tex]² - v[tex]_i[/tex]² )
ocean currents are always cold true or false
what makes a funnel appear black
Answer:
Air
Explanation:
The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.
If
[tex] \binom{13}{7} [/tex]
N has a half-life of about 10.0 min, how long will it take for 20 g of the isotope to decay to 1.9 g? [2 marks] 2. Radon-22
Answer:
the time it will take the element to decay to 1.9 g is 34.8 mins.
Explanation:
Given;
half life of Nitrogen, t = 10 min
initial mass of the element, N₀ = 20 g
final mass of the element, N = 1.9 g
The time taken for the element to decay to final mass is calculated as follows;
time (min) mass remaining
0 ----------------------------------20 g
10 mins ------------------------- 10 g
20 mins ------------------------- 5 g
30 mins -------------------------- 2.5 g
40 mins --------------------------- 1.25 g
Interpolate between 2.5 g and 1.25 to obtain the time for 1.9 g
30 min ------------------------- 2.5 g
x ----------------------------------- 1.9 g
40 min -------------------------- 1.25 g
[tex]\frac{30 - x}{40- 30} = \frac{2.5 - 1.9}{1.25 - 2.5} \\\\-1.25(30-x) = 6\\\\-37.5 + 1.25x = 6\\\\1.25x = 6+37.5\\\\1.25x = 43.5\\\\x = \frac{43.5}{1.25} \\\\x = 34.8 \ mins[/tex]
Therefore, the time it will take the element to decay to 1.9 g is 34.8 mins.
A 0.100 kg limestone cube is released from rest, and proceeds to slide down a frictionless ramp. At the bottom of the ramp, the limestone cube makes an elastic collision with a steel cube whose mass is 0.200 kg, which is initially at rest. At what vertical height should the limestone cube be placed such that the steel cube has a velocity of 1.50 m/s after the collision
Answer:
the height at which the limestone cube must be placed is 0.23 m.
Explanation:
Given;
mass of limestone cube, m₁ = 0.1 kg
initial velocity of the limestone cube, u₁ = 0
mass of steel cube, m₂ = 0.2 kg
initial velocity of the steel cube, u₂ = 0
final velocity of the steel cube, v₂ = 1.5 m/s
Apply the principle of conservation of energy to determine the height of the limestone cube.
Potential energy of the limestone cube at top = Kinetic energy of steel cube at base
m₁gh = ¹/₂m₂v₂²
where;
h is the height at which the limestone cube is placed
[tex]h = \frac{m_2v_2^2}{2m_1g} \\\\h = \frac{0.2\times 1.5 ^2}{2 \times 0.1 \times 9.8} \\\\h =0.23 \ m[/tex]
Therefore, the height at which the limestone cube must be placed is 0.23 m.
When a wave enters a new medium from an angle, both the speed and the ________ change
a
The frequency
b
The amplitude
c
The energy
d
The angle
Answer:
B: Amplitude
Explanation:
When a wave travels from one medium to the other from an angle, the things that change are amplitude, wavelength, intensity and velocity.
The frequency doesn't change because the frequency depends upon the source of the wave and not the medium by which the wave is propagated.
Answer:
The angle
Explanation:
Define the average acceleration of a particle
between two given instants.
help me with the question in the image I provided
Answer: 10 s, 30 m/s , 150 m
Explanation:
Given
The speed of motorcyclist is [tex]u_1=15\ m/s[/tex]
The initial speed of a police motorcycle is [tex]u_2=0\ m/s[/tex]
acceleration of police motorcycle is [tex]a=3\ m/s^2[/tex]
Police will catch the motorcyclist when they traveled equal distances
distance traveled by motorcyclist in time t is
[tex]\Rightarrow s_1=15\times t[/tex]
Distance traveled by Police in time t is
[tex]\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2[/tex]
put [tex]s_1=s_2[/tex]
[tex]\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s[/tex]
Police officer's speed at that time is
[tex]\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s[/tex]
Distance traveled by each vehicle is
[tex]\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m[/tex]
When sound waves travels from air to water
Answer:
Frequency of a sound wave remains independent of the medium through which it travels. Wavelength, velocity and phase change with the change in media.
Explanation:
Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M and radius R. The axis of rotation passes through the center of each object, and is perpendicular to the plane of the hoop and the plane of the flat disk. Which of these objects requires the largest torque to give it the same angular acceleration?
a. the solid sphere.
b. the hollow sphere.
c. the hoop.
d. the flat disk.
e. both the solid and the hollow spheres.
Answer:
b) The hollow sphere.
Explanation:
The moment of Inertia of a solid rotating object is related with the torque applied and the angular acceleration caused by the torque, by the following relationship, that resembles Newton's 2nd Law for point masses:[tex]\tau = I * \alpha (1)[/tex]
As it can be seen, for a given angular acceleration α, the larger the moment of inertia, the larger the torque needed to give the object the same angular acceleration.From the proposed solids, the one that has the largest moment of Inertia, is the hollow sphere, which moment of Inertia can be written as follows:[tex]I_{hsph} = \frac{2}{3} *M*R^{2} (2)[/tex]
So, the hollow sphere requires the largest torque to give the object the same angular acceleration.Suppose that you changed the area of the bottom surface of the friction cart without changing its mass, by replacing the Teflon slab with one that was smaller but thicker. The contact area would shrink, but the normal force would be the same as before. Would this change the friction force on the sliding cart
Answer:
in this case the weight of the vehicle does not change , consequently the friction force should not change
Explanation:
The friction force is a macroscopic manifestation of the interactions of the molecules between the two surfaces, this force in the case of solid is expressed by the relation
fr = μ N
W-N= 0
N = W
as in this case the weight of the vehicle does not change nor does the Normal one, consequently the friction force should not change
Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2
A 0.86kg grenade is tossed on the ground with a velocity of 6 m/s West. If the grenade explodes into 2 pieces,
one that has a mass of 32kg and travels East at 10 m/s.
(A) What is the mass of the second piece
(B) What is the velocity of the second piece?
Answer:
(A) 0.54 kg
(B) 15.5 m/s west
Explanation:
Mass is conserved.
M = m₁ + m₂
0.86 kg = 0.32 kg + m₂
m₂ = 0.54 kg
Momentum is conserved (take east to be positive).
Mv = m₁v₁ + m₂v₂
(0.86 kg)(-6 m/s) = (0.32 kg)(10 m/s) + (0.54 kg) v₂
v₂ = -15.5 m/s
. If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder
Answer:
As we can see, a string is attached with block A, and three string is folded with ply which is attached with B
x
B
=3x
A
Now differentiate with respect to x
V
B
=3V
A
Given,
V
A
=0.6m/s(totheright)
So,
V
B
=0.6×3
=1.8m/s(downward)
Explanation:
IF THE ANSWER IS RIGHT PLZ GIVE ME BRAINLIEST
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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.2 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.3 mm and that is perpendicular to the direction of the field.
poste en français s’il vous plaît
Given equal acceleration, which has
most force?
Mosquito
Dog
Horse
Elephant
Object A is negatively charged. Object A and Object B
attract. Object B and Object C repel. Object C and Object
D repel. What type of charge does Object B, Object C, and
Object D possess?
Answer:
Malrpr00qpq9owoowopwiaahaulaqkkkala9asoLHahababajjajalls
Explanation:
hhoootyiñlf7ogffyiklmhf
In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find
an electron's acceleration in this field.
Answer:
a = 1.055 x 10¹⁷ m/s²
Explanation:
First, we will find the force on electron:
[tex]E = \frac{F}{q}\\\\F = Eq\\[/tex]
where,
F = Force = ?
E = Electric Field = 6 x 10⁵ N/C
q = charge on electron = 1.6 x 10⁻¹⁹ C
Therefore,
[tex]F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\[/tex]
F = 9.6 x 10⁻¹⁴ N
Now, we will calculate the acceleration using Newton's Second Law:
[tex]F = ma\\a = \frac{F}{m}\\[/tex]
where,
a = acceleration = ?
m = mass of electron = 9.1 x 10⁻³¹ kg
therefore,
[tex]a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\[/tex]
a = 1.055 x 10¹⁷ m/s²
If a true bearing of a ship at sea is 227°, what is its direction angle?
A. 43°
B.313°
C. 223°
A piston-cylinder device initially contains 0.6 kg of water with a volume of 0.1 m3 . The mass of the piston is such that it maintains a constant pressure of 1000 kPa. The cylinder is connected through a valve to a supply line that carries steam at 5 MPa and 500 o C. Now the valve is opened and steam is allowed to flow slowly into the cylinder until the volume of the cylinder doubles and the temperature in the cylinder reaches 280 o C, at which point the valve is closed. The pressure remains constant during the process. Determine:
Answer: Hello the missing piece of your question is attached
question : Determine mass of steam that has entered ( in kg )
answer : 0.206 kg
Explanation:
V1 = 0.1 m^3 ,
v' = V1 / m1 = 0.1 / 0.6 = 0.167 m^3/kg
V2 = 0.2 m^3
using the steam tables
at ; P = 1000 kPa, v' = 0.167 m^3/kg
U1 = 2321 KJ/kg
at ; P = 1000 kPa , T2 = 280°C
v'2= 0.2481 m^3kg
U2 = 2760.6
at ; P = 5MPa , T = 500°C
h1 = 3434.7 KJ/Kg
calculate final mass ( m2 )
M2 = V2 / v'2
= 0.2 / 0.2481 = 0.806 kg
therefore the mass added = m2 - m1
= 0.806 - 0.6 = 0.206 kg
A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow
Answer:
132
Explanation:
if you round it is correct
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Calculate the amount of torque of an object being pushed by 6 N force along a circular path of a radius of 1x10^-2 mat 30 degree angle
Answer:
[tex]\tau=0.03\ N-m[/tex]
Explanation:
Given that,
Force acting, F = 6N
The radius of the path, [tex]r=10^{-2}\ m[/tex]
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find the amount of torque acting on the object. The formula for torque is given by :
[tex]\tau=Fr\sin\theta\\\\\tau=6\times 10^{-2}\times \sin(30)\\\\\tau=0.03\ N-m[/tex]
So, the required torque is equal to 0.03 N-m.
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
PLEASE I NEED HELP CLICK ON THIS IMAGE
The circuit is working, and all three bulbs are lit. If a switch at B is opened, what will happen to the circuit? Photo included PLS ONLY ANSWER IF YOU KNOW 100%
Answer:
D. Bulb 2 will go out but bulbs 1 and 3 will remain lit.
Explanation:
If switch at B is opened then, Bulb 2 will go out but bulbs 1 and 3 will remain lit. This is because the positive and negative current flow still fully passes through the two closed switches at switch A and C which allows the bulbs to receive both currents and thus allowing it to turn on. When Switch B is opened it cuts off access for the negative charge to get to bulb 2 and without it, bulb 2 will not turn on.
Answer:all 3 bulbs will go out
Explanation:
Just did it and got it right
What are the two main processes carried out by the excretory system?
3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is
swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?
Answer:
the velocity of the big fish after the launch is 6 m/s.
Explanation:
Given;
mass of the big fish, m₁ = 5 kg
velocity of the big fish, u₁ = 8 m/s
mass of the small fish, m₂ = 1 kg
velocity of the small fish, u₂ = -4 m/s
Let the final velocity of the big fish after launch = v
Apply the principle of conservation linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
5 x 8 + 1 x (-4) = v(5 + 1)
40 - 4 = 6v
36 = 6v
v = 36/6
v = 6 m/s.
Therefore, the velocity of the big fish after the launch is 6 m/s.
how many atmospheres is at depth of 100 meters of ocean water?
Explanation:
At 100m above sea level, the air pressure is approximately 990mbar, so the air pressure has decreased by approximately 10mbar.
There would be 10 atmospheric pressure at a depth of 100 meters of ocean water because approximately 10 meters depth of water is equivalent to 1 atmospheric pressure.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
As given in the problem we have to find out the atmospheric pressure equivalent at a depth of 100 meters of ocean water,
10 meters depth of water = 1 atmospheric pressure
100-meter depth of the ocean water = 10 atmospheric pressure
Thus, there would be 10 atmospheric pressure at a depth of 100 meters of ocean water.
To learn more about pressure, refer to the link given below ;
brainly.com/question/28012687
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effect of high pitch on humans
Answer:
High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.