Ask question

Ask question

All categories

2 months ago

12

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration ofa= 2.60 m/s^2. A worker assists t

he cart by pushing on the crate with a force that is eastward andhas magnitude that depends on time. According to F(t) = (5.40 N/s)t. What is the instantaneouspower supplied by this force at t= 4.70 s

1 answer:

Keith_Richards [3.2K]2 months ago

8 0

Response:

To find power, we must first determine the work done by the force.

1) We will use the following equation to calculate work:

$\int\limits {F} \, dx$

The force is provided by the problem; our goal is to express 'dx' in terms of 't'

2) It's known that:

$\frac{dV}{dt} = a = 2.6$

Thus, we have:

$v = 2.6t$

Then:

$\frac{dx}{dt} = V = 2.6t\\ \\dx = 2.6t*dt$

3) Finally, substituting all known values gives us:

$\int\limits^{4.7}_{0} {5.4t*2.6t} \, dt$

After some calculations, the resulting work is:

161.9638 J.

4) To find power, we will use the following equation:

$P = \frac{W}{t}$

Thus

P = 161.9638/4.7 = 34.46 W

You might be interested in

1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What

Ostrovityanka [3204]

Response:

2.5kN.m

Details:

Torque relates directly to the pitch diameter

= Ta/Tb= Da/Db

For 120/Tb= 0.25/0.5

This gives Tb= 2.469kN.m, roughly 2.5kN.m

8 0

2 months ago

Read 2 more answers

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Ostrovityanka [3204]

Answer: The calorimeter's heat capacity is $6.72J/g^oC$

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of the mixture = $35.0^oC$

$T_1$ = initial temperature of the water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Hence, the heat capacity of the calorimeter is $6.72J/g^oC$

3 0

3 months ago

What is the internal energy (to the nearest joule) of 10 moles of Oxygen at 100 K?

serg [3582]

Response:

U = 12,205.5 J

Clarification:

To determine the internal energy of an ideal gas, use the following equation:

$U=\frac{3}{2}nRT$ (1)

U: internal energy

R: ideal gas constant = 8.135 J(mol.K)

n: number of moles = 10 mol

T: the temperature of the gas = 100K

Substituting the parameter values into equation (1):

$U=\frac{3}{2}(10mol)(8.135\frac{J}{mol.K})(100K)=12,205.5J$

The overall internal energy for 10 moles of Oxygen at 100K is 12,205.5 J

6 0

1 month ago