The uniform crate has a mass of 50 kg and rests on the cart having an inclined surface. Determine the smallest acceleration that will cause the crate either to tip or slip relative to the cart. What is the magnitude of this acceleration

Answers

Answer 1
Answer:

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:

50asin(15)x+50acos(15)0.5=-50(9.81)sin(15)0.5+50(9.81)cos(15)x\ \ \ (1)

Sum of vertical forces is zero, hence:

N-50(9.81)cos(15)+50acos(15)=0\ \ \ (2)

Sum of horizontal force is zero, hence:

50(9.81)sin(15)-\mu N+50acos(15)=0\n\n50(9.81)sin(15)-0.5 N+50acos(15)=0\ \ \ (3)

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


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An advantage of J.J. Thomson's Plum Pudding Model was that it _____. A. was a much less expensive way to study atoms B. simplified the calculations necessary to describe an atomC. clearly explained where electrons were located in an atomD. is much less expensive to bake a plum pudding than to look at an atom

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

Answers

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

Final answer:

The second hole should be cut at the same height as the first hole to have the same range for the stream.

Explanation:

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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Why a switch is connected in phase wire and never is neutral wire?​

Answers

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

Answer:

ifyou have a wre connect you should not have to connected

i think that is the answer

Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal energy dissipated from brakes in a 1600 kg car that descends a 15 ∘ hill. The car begins braking when its speed is 95 km/h and slows to a speed of 40 km/h in a distance of 0.34 km measured along the road.

Answers

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.

Final answer:

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.

Explanation:

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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"Which gives the transverse acceleration of an element on a string as a wave moves along an x axis along the string?"

Answers

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

Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?

Answers

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,

f_(b)=f -(√(T * f))

Solving as,

f_(b)=294-(√(0.98 * 294))\n\nf_(b)=2.95\;\rm Hz

Thus, we can conclude that the beat frequency heard when the two strings are played together is 2.95 Hz.

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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)

Write down the DE of motion of a particle moving under the influence of gravity and experiencing a resistive force. .

Answers

Answer:

The DE will be (d^2x)/(dt^2)-(k)/(m)(dx)/(dt)-g=0

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 =mg

Le the a resistive force of magnitude kv opposes this gravity force, here k is constant and v is velocity.

So net force F_(NET)=mg-kv-----eqn 1

F_(NET)=ma

So ma=mg-kv

We know that velocity is rate of change of position so v=(dx)/(dt), and acceleration is rate of change of velocity so a=(d^2x)/(dt^2)

Putting all these value in eqn 1

m(d^2x)/(dt^2)=mg-k(dx)/(dt)

(d^2x)/(dt^2)-(k)/(m)(dx)/(dt)-g=0