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Mazyrski [523]
2 years 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
Physics
1 answer:
Leona [35]2 years ago
8 0

Answer:

To obtain the power, we first need to find the work made by the force.

1) To calculate the work, we need the next equation:

\int\limits {F} \, dx

So the force is given by the problem so our mission is to find 'dx' in terms of 't'

2) we know that:

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

So we have:

v = 2.6t

Then:

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

3) Finally, we replace everything:

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

After some calculation, we have as a result that the work is:

161.9638 J.

4) To calculate the power we need the next equation:

P = \frac{W}{t}

So

P = 161.9638/4.7 = 34.46 W

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The magnetising current in a transformer is rich in ______________ harmonic?
sukhopar [10]
The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher. 
5 0
3 years ago
Earth exerts a 100 N gravitational force on a metal box. What is the magnitude of the gravitational force the metal box exerts o
jeka94

We have that the magnitude of the gravitational force  is mathematically given as

f=6.377N

<h3>Force</h3>

Question Parameters:

Earth exerts a 100 N gravitational force on a metal box.

(Mass of the earth is 6e24 kg and radius of the earth is 6.4e6m.)

Generally the equation for the Gravitational mForce  is mathematically given as

F=\frac{GMm}{r^2}\\\\Therefore\\\\F=\frac{100/9.8* 6e116e24}{6.4e6^2}

f=6.377N

For more information on Force visit

brainly.com/question/26115859

7 0
2 years ago
A nylon guitar string is fixed between two lab posts 2.00 m apart. The string has a linear mass density of μ=7.20 g/m\mu=7.20~\t
vladimir2022 [97]

Answer:

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

\mu =7.20 g/m = 0.0072 kg/m is the mass linear density

Solving the equation,

f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

f=\frac{v}{2L}

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m

4 0
3 years ago
3. Identify three elements and three compounds. How are they<br> similar? How are they different?
Basile [38]

Answer:  A compound is a substance formed when two or more elements are chemically joined. Water, salt, and sugar are examples of compounds. When the elements are joined, the atoms lose their individual properties and have different properties from the elements they are composed of.

6 0
3 years ago
A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
viva [34]

Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

5 0
3 years ago
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