Answer:
The answer is below
Explanation:
Let g = acceleration due to gravity = 9.81 m/s², x = half of the width of the crate, half of the height of the crate = 0.5 m, a = acceleration of crate, N = force raising the crate
The sum of moment is given as:
Sum of vertical forces is zero, hence:
Sum of horizontal force is zero, hence:
Solving equation 1, 2 and 3 simultaneously gives :
N = 447.8 N, a = 2.01 m/s², x = 0.25 m
x is supposed to be 0.3 m (0.6/2)
The crate would slip because x <0.3 m
Answer: hello your question is incomplete below is the complete question
Water stands at a depth H in a large open tank whose side walls are vertical . A hole is made in one of the walls at a depth h below the water surface. Part B How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole
answer :
At Height ( h ) from the bottom of Tank
Explanation:
Determine how far above the bottom of the tank a second hole be cut
For the second hole to have the same range as the first hole
Range of first hole = Velocity of efflux of water * time of fall of water
= √ (2gh) * √( 2g (H - h) / g)
= √ ( 4(H-h) h)
Hence the Height at which the second hole should be placed to exercise same range of stream emerging = h from the bottom of the Tank
The second hole should be cut at the same height as the first hole to have the same range for the stream.
In order for the stream emerging from the second hole to have the same range as the first hole, the second hole should be cut at the same height as the first hole. This is because the range of the stream depends on the initial velocity and the vertical distance traveled. If the second hole is higher or lower than the first hole, the vertical distance traveled will be different and the range of the stream will be affected.
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Answer:
ifyou have a wre connect you should not have to connected
i think that is the answer
Answer:
1838216 J
Explanation:
95 km/h = 26.39 m/s
40 km/h = 11.11 m/s
Initial kinetic energy
= .5 x 1600 x(26.39)²
= 557145.67 J
Final kinetic energy
= .5 x 1600 x ( 11.11)²
= 98745.68 J
Loss of kinetic energy
= 458400 J
Loss of potential energy
= mg x loss of height
= 1600 x 9.8 x 340 sin 15
= 1379816 J
Sum of Loss of potential energy and Loss of kinetic energy
= 1379816 + 458400
= 1838216 J
This is the work done by the friction . So this is heat generated.
To calculate the thermal energy dissipated from the brakes of a car, use the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill. The temperature change of the brakes can then be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.
The thermal energy dissipated from the brakes of a car can be calculated by converting the gravitational potential energy lost by the car into internal energy of the brakes. By using the equation Q = Mgh/10, where Q is the energy transferred to the brakes, M is the mass of the car, g is the acceleration due to gravity, and h is the height of the hill, we can calculate the thermal energy dissipated. From there, the temperature change of the brakes can be calculated using the equation Q = mc∆T, where m is the mass of the brakes and c is its specific heat capacity.
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Answer:
the second derivative of y with respect to time gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string
Explanation:
This is because the transverse wave movement of particles take place in direction 90° to direction of movement of the wave (x) itself, so second derivative of y with respect to time (t)is what will be required
The beat frequency heard when the two strings are played together is 2.95 Hz.
Given data:
The tuning frequency of the violin is, f = 294 Hz.
Decrement in the tension is, 2 %.
Since, tension is reduced at the rate of 2%. Then the new magnitude of tension on the string is,
T = (100 - 2 )/100
T = 0.98
Then the expression for the beat frequency heard when the two strings are played together is given as,
Solving as,
Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.
Learn more about the beat frequency here:
Answer:
Beat frequency together = 2.95 Hz (Approx)
Explanation:
Given:
Frequency (F) = 294 H
Decrease in tension = 2%
Find:
Beat frequency together
Computation:
Tension = (100 - 2) / 100
Tension (T) = 0.98
Beat frequency together = Frequency (F) - (√T × F)
Beat frequency together = 294 - (√0.98 × 294)
Beat frequency together = 2.95 Hz (Approx)
Answer:
The DE will be
Explanation:
We have to find differential equation under the influence of gravity and experiencing a resistive force
Let an object of mass m falling under the influence of gravity
So the force experience under gravity
Le the a resistive force of magnitude kv opposes this gravity force, here k is constant and v is velocity.
So net force -----eqn 1
So
We know that velocity is rate of change of position so , and acceleration is rate of change of velocity so
Putting all these value in eqn 1