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3 years ago

How long does it take a racehorse to accelerate from rest to a speed of 15 m/s with an acceleration of 3.0 m/s2?

Physics

2 answers:

GuDViN [60]3 years ago

8 0

It would take 5.0 s for a racehorse to accelerate from rest

aleksandr82 [10.1K]3 years ago

4 0

It would take 5.0 s for a racehorse to accelerate from rest to a speed of 15 m/s with an acceleration of 3.0 m/s.

Take 15 / 3.0 to get your answer of D. 5.0 s

Hope that helps u out!

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At which point or points does the pendulum have the greatest kinetic energy?

Daniel [21]

At the lowest point of its motion, kinetic energy is maximum and potential energy is minimum. This is where the velocity is a maximum. At the highest point of its motion, kinetic energy is minimum (i.e. zero) and potential energy is maximum.

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3 years ago

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Steam is to be condensed on the shell side of a heat exchanger at 150 oF. Cooling water enters the tubes at 60 oF at a rate of 4

zalisa [80]

Answer:

a. 572Btu/s

b.0.1483Btu/s.R

Explanation:

a.Assume a steady state operation, KE and PE are both neglected and fluids properties are constant.

From table A-3E, the specific heat of water is $c_p=1.0\ Btu/lbm.F$ , and the steam properties as, A-4E:

$h_{fg}=1007.8Btu/lbm, s_{fg}=1.6529Btu/lbm.R$

Using the energy balance for the system:

$\dot E_{in}-\dot E_{out}=\bigtriangleup \dot E_{sys}=0\\\\\dot E_{in}=\dot E_{out}\\\\\dot Q_{in}+\dot m_{cw}h_1=\dot m_{cw}h_2\\\\\dot Q_{in}=\dot m_{cw}c_p(T_{out}-T_{in})\\\\\dot Q_{in}=44\times 1.0\times (73-60)=572\ Btu/s$

Hence, the rate of heat transfer in the heat exchanger is 572Btu/s

b. Heat gained by the water is equal to the heat lost by the condensing steam.

-The rate of steam condensation is expressed as:

$\dot m_{steam}=\frac{\dot Q}{h_{fg}}\\\\\dot m_{steam}=\frac{572}{1007.8}=0.5676lbm/s$

Entropy generation in the heat exchanger could be defined using the entropy balance on the system:

$\dot S_{in}-\dot S_{out}+\dot S_{gen}=\bigtriangleup \dot S_{sys}\\\\\dot m_1s_1+\dot m_3s_3-\dot m_2s_2-\dot m_4s_4+\dot S_{gen}=0\\\\\dot m_ws_1+\dot m_ss_3-\dot m_ws_2-\dot m_ss_4+\dot S_{gen}=0\\\\\dot S_{gen}=\dot m_w(s_2-s_1)+\dot m_s(s_4-s_3)\\\\\dot S_{gen}=\dot m c_p \ In(\frac{T_2}{T_1})-\dot m_ss_{fg}\\\\\\\dot S_{gen}=4.4\times 1.0\times \ In( {73+460)/(60+460)}-0.5676\times 1.6529\\\\=0.1483\ Btu/s.R$

Hence,the rate of entropy generation in the heat exchanger. is 0.1483Btu/s.R

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3 years ago

A transform boundary occurs where two tectonic plates _____.?

gladu [14]

Where they slide over each other.

Transform boundaries are formed or occur when two plates slide past each other in a sideways motion. They do not tear or crunch into each other (but the rock in between them may be ground up) and therefore none of the spectacular features are seen such as occur in divergent and convergent boundaries.

In a transform boundary, neither plate is added to at the boundary nor destroyed. They are marked in some places by features like stream beds that have been split in half and the two halves moved in opposite directions.

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4 years ago

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Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$