Ask question

Ask question

All categories

2 years ago

12

A hockey puck is pushed by a stick with a force of 750 newtons. The puck travels 2.0 meters in 0.30 seconds. How powerful is the

push? (Ignore frictional effects.)

1 answer:

nekit [7.7K]2 years ago

5 0

The work done by the force pushing the puck is equal to the product between the force and the distance covered:
$W=Fd=(750 N)(2.0 m)=1500 J$

And the power is the work done per unit time:
$P= \frac{W}{t}$
and using t=0.30 s, we find the power:
$P= \frac{1500 J}{0.30 s}=5000 W$

You might be interested in

A jetboat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, start fraction, start text, m, end tex

love history [14]

The question is incomplete. Here is the entire question.

A jetboat is drifting with a speed of 5.0m/s when the driver turns on the motor. The motor runs for 6.0s causing a constant leftward acceleration of magnitude 4.0m/s². What is the displacement of the boat over the 6.0 seconds time interval?

Answer: Δx = - 42m

Explanation: The jetboat is moving with an acceleration during the time interval, so it is a <u>linear</u> <u>motion</u> <u>with</u> <u>constant</u> <u>acceleration</u>.

For this "type" of motion, displacement (Δx) can be determined by:

$\Delta x = v_{i}.t + \frac{a}{2}.t^{2}$

$v_{i}$ is the initial velocity

a is acceleration and can be positive or negative, according to the referential.

For Referential, let's assume rightward is positive.

Calculating displacement:

$\Delta x = 5(6) - \frac{4}{2}.6^{2}$

$\Delta x = 30 - 2.36$

$\Delta x$ = - 42

Displacement of the boat for t=6.0s interval is $\Delta x$ = - 42m, i.e., 42 m to the left.

8 0

2 years ago

he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold

bekas [8.4K]

Answer:

Relative population is 2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-E_{1} }{KT} } }{g_{2}e^{\frac{-E_{2} }{KT} }}$

$\frac{N_{1} }{N_{2} }$ = $\frac{g_{1}e^{\frac{-(E_{1}-E_{2}) }{KT} } }{g_{2}}$

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

$\frac{N_{1} }{N_{2} }$ = $\frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} } }{3}$

$\frac{N_{1} }{N_{2} }$ = 3.4 x 10⁹

$\frac{N_{2} }{N_{1} }$ = 2.94 x 10⁻¹⁰

5 0

2 years ago

A block spring system oscillates on a frictionless surface with an amplitude of 10\text{ cm}10 cm and has an energy of 2.5 \text

antoniya [11.8K]

Answer:

The energy of the system is 15 J.

Explanation:

Given that,

Energy E = 2.5 J

Amplitude = 10 cm

We need to calculate the spring constant

Using formula of mechanical energy of the system

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$2.5=\dfrac{1}{2}k\times(10\times10^{-2})^2$

$k=\dfrac{2.5\times2}{(10\times10^{-2})^2}$

$k=500\ N/m$

If the block is replaced by a block with twice the mass of the original block

Amplitude = 6 cm

We need to calculate the energy

Using formula of mechanical energy

$E=\dfrac{1}{2}kA^2$

Put the value into the formula

$E=\dfrac{1}{2}\times500\times(6\times10^{-2})$

$E=15\ J$

Hence, The energy of the system is 15 J.

8 0

2 years ago

A 110-pound person pulls herself up 4.0 feet. This took her 2.5 seconds. How much power was developed?

mrs_skeptik [129]

Answer:

0.32 Hp is the answer. right

5 0

1 year ago

Read 2 more answers

An insurance company hired your group to help investigate an insurance claim following a car accident. In the accident, two cars

Romashka-Z-Leto [24]

Answer:

v=8m/s

Explanation:

To solve this problem we have to take into account, that the work done by the friction force, after the collision must equal the kinetic energy of both two cars just after the collision. Hence we have

$W_{f}=E_{k}\\W_{f}=\mu N=\mu(m_1+m_1)g\\E_{k}=\frac{1}{2}[m_1+m_2]v^2$

where

mu: coefficient of kinetic friction

g: gravitational acceleration

We can calculate the speed of the cars after the collision by using

$W_f=(0.7)(1650kg+1900kg)(9.8\frac{m}{s^2})=24353J\\24353J=\frac{1}{2}(1650kg+1900kg)v^2\\v=\sqrt{\frac{24353J}{1775kg}}=3.70\frac{m}{s}$

Now , we can compute the speed of the second car by taking into account the conservation of the momentum

$P_b=P_a\\m_1v_1+m_2v_2=(m_1+m_2)v\\\\v_2=\frac{(m_1+m_2)v-m_1v_1}{m_1}\\\\v_2=\frac{(1650kg+1900kg)(3.7\frac{m}{s})-(1650kg)(16\frac{m}{s})}{(1650kg)}\\\\v_2=8\frac{m}{s}$

the car did not exceed the speed limit

Hope this helps!!

7 0

2 years ago

Read 2 more answers