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4 years ago

5

A hammer slides down a roof that makes a 30.0°angle with the horizontal. What are the magnitudes of the components of the hammer

's velocity at the edge of the roof if it is moving at a speed of 8.25 m/s

1 answer:

dlinn [17]4 years ago

7 0

<h2>Answer:7.14 $ms^{-1}$ ,4.125 $ms^{-1}$ </h2>

Explanation:

Whenever an object is moving in a 2D frame,its motion can be analysed as if it is travelling in two independent 1D frames.

One of such independent 1D frames are along horizontal and another along vertical.

Let $v$ be the total velocity.

Given that, $v=8.25ms^{-1}$

We call the horizontal velocity as $v_{h}$ and the vertical velocity as $v_{v}$ .

$v_{h}$ = $vCos\alpha$

$v_{v}=vSin\alpha$

where $\alpha$ is the angle between the object and horizontal.

It is given that $\alpha =30^{0}$

$v_{h}=8.25\times Cos(30^{0})=7.14ms^{-1}$

$v_{v}=8.25\times Sin(30^{0})=4.125ms^{-1}$

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can you suggest improvement that can be made towards the design of siphon so that the transfer of liquid is much higher.

ahrayia [7]

Answer:

A siphon is a tube that makes use of the potential energy of fluid at an elevated level to transfer the fluid to a lower level, due to pressure differences between the inlet and the outlet points of the tube, such that the pressure at the outlet is higher than the pressure at the inlet

The pressure energy is converted into velocity (kinetic) energy, and therefore, in other to increase the flow rate through the tube of a siphon, with constant diameter, the level of the fluid in the container at the inlet (supply) of the siphon is raised higher than the level at the outlet receiving) container or the outlet point of the siphon tube

The larger the difference between the inlet and outlet levels, the faster the transfer of fluid by the siphon

Explanation:

3 0

3 years ago

If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel

expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

$V = u + at$

Substituting into the equation, we have;

$V = 0 + 9.8 * 3.6$

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

$S = ut + \frac {1}{2}at^{2}$

Substituting the values into the equation;

$S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}$

$S = 0 + 4.9*12.86$

$S = 0.5 *36$

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0

3 years ago

According to the Doppler affect what happens when a light surfaceMoves further away from an observer

Naily [24]

The correct answer is The electromagnetic waves appear more in red color.
<span>Since red is at the low-frequency end of the visible spectrum, we say that light from a receding star is shifted toward red, or redshifted.</span>

8 0

3 years ago

Read 2 more answers

Having difficulty finding the PE and KE for these values no mass is given. Does anyone know to go solve these?

Alexandra [31]

11) $1.04\cdot 10^7 J$

12) $1.04\cdot 10^7 J$

13) 50.0 m/s

14) 41.6 m/s

Explanation:

11)

The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by

$PE=mgh$

where

m is the mass of the object

g is the acceleration due to gravity

h is the height relative to the ground

Here in this problem, when the train is at the top, we have:

m = 8325 kg (mass of the train + riders)

$g=9.8 m/s^2$ (acceleration due to gravity)

h = 127 m (height of the train at the top)

Substituting,

$PE=(8325)(9.8)(127)=1.04\cdot 10^7 J$

12)

According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:

$KE_t + PE_t = KE_b + PE_b$

where

$KE_t$ is the kinetic energy at the top

$PE_t$ is the potential energy at the top

$KE_b$ is the kinetic energy at the bottom

$PE_b$ is the potential energy at the bottom

The kinetic energy is the energy due to motion; since the train is at rest at the top, we have

$KE_t=0$

Also, at the bottom the height is zero, so the potential energy is zero

$PE_b=0$

Therefore, we find:

$KE_b=PE_t=1.04\cdot 10^7 J$

13)

The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by

$KE=\frac{1}{2}mv^2$

where

m is the mass of the object

v is the speed of the object

From question 12), we know that the kinetic energy of the train at the bottom is

$KE=1.04\cdot 10^7 J$

We also know that the mass is

m = 8325 kg

Therefore, we can calculate the speed of the train at the bottom:

$v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(1.04\cdot 10^7)}{8325}}=50.0 m/s$

14)

At the top of the second hill, the total mechanical energy of the train is still conserved.

Therefore, we can write again:

$KE_1 + PE_1 = KE_2 + PE_2$

where

$KE_1$ is the kinetic energy at the top of the 1st hill

$PE_1$ is the potential energy at the top of the 1st hill

$KE_2$ is the kinetic energy at the top of the 2nd hill

$PE_2$ is the potential energy at the top of the 2nd hill

From the previous questions, we know that

$KE_1=0$

and

$PE_1=1.04\cdot 10^7 J$

The height of the second hill is

h = 39 m

So we can also find the potential energy at the second hill:

$PE_2=mgh=(8325)(9.8)(39)=3.2\cdot 10^6 J$

So, the kinetic energy at the second hill is

$KE_2=PE_1-PE_2=1.04\cdot 10^7 - 3.2\cdot 10^6 =7.2\cdot 10^6 J$

And so, the speed is

$v=\sqrt{\frac{2KE_2}{m}}=\sqrt{\frac{2(7.2\cdot 10^6)}{8325}}=41.6 m/s$

4 0

3 years ago

A toy cannon uses a spring to project a 5.24-g soft rubber ball. The spring is originally compressed by 5.01 cm and has a force

salantis [7]

Answer:

Speed will be equal to 1.40 m/sec

Explanation:

Mass of the rubber ball m = 5.24 kg = 0.00524 kg

Spring is compressed by 5.01 cm

So x = 5.01 cm = 0.0501 m

Spring constant k = 8.08 N/m

Frictional force f = 0.031 N

Distance moved by ball d = 15.8 cm = 0.158 m

Energy gained by spring

$KE=\frac{1}{2}kx^2=\frac{1}{2}\times 8.08\times 0.0501^2=0.0101J$

Energy lost due to friction

$W=Fd=0.031\times 0.158=0.0048J$

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J

This energy will be kinetic energy

$\frac{1}{2}mv^2=0.0052$

$\frac{1}{2}\times 0.00524\times v^2=0.0052$

v = 1.40 m/sec

7 0

3 years ago