Ask question

Ask question

All categories

2 years ago

9

Two 0.5 kg carts,one red and one green ,sits about half a meter apart on a low friction track. You push on the red with a consta

nt force of 4N for 0.17m and then remove your hand. The cart moves 0.33 m on the track and then strikes the green cart. What is the work done by you on the two cart system?

1 answer:

xxMikexx [17]2 years ago

6 0

Answer:

Work done on the two cart system is 2 N-m

Explanation:

Work done is equal to the product of force and displacement

Work done =4N * (0.17 +0.33)

work done = 4N * 0.5

Work done = 2 N -m

Work done on the two cart system is 2 N-m

You might be interested in

What type of change accurs when a substance stays the same

Thepotemich [5.8K]

This would be a physical change because it can change back to its original form. This is like ripping paper. You can piece it back together and it still is paper.

The opposite of this is chemical change. Chemical change means the product has been changed completely like burning paper. The paper has now been turned to ash and it's impossible to change this back to its original form.

3 0

3 years ago

A beam of light strikes a sheet of glass at an angle of 56.6° with the normal in air. You observe that red light makes an angle

Yuri [45]

Answer:

(a). Index of refraction are $n_{red}$ = 1.344 & $n_{violet}$ = 1.406

(b). The velocity of red light in the glass $v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

The velocity of violet light in the glass $v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

Explanation:

We know that

Law of reflection is

$n_1 \sin\theta_{1} = n_2 \sin\theta_{2}$

Here

$\theta_1$ = angle of incidence

$\theta_2$ = angle of refraction

(a). For red light

1 × $\sin 56.6$ = $n_{red}$ × $\sin 38.4$

$n_{red}$ = 1.344

For violet light

1 × $\sin 56.6$ = $n_{violet}$ × $\sin 36.4$

$n_{violet}$ = 1.406

(b). Index of refraction is given by

$n = \frac{c}{v}$

$n_{red}$ = 1.344

$v_{red} = \frac{c}{n_{red} }$

$v_{red} = \frac{3(10^{8} )}{1.344}$

$v_{red} =$ 2.23 × $10^{8} \ \frac{m}{s}$

This is the velocity of red light in the glass.

The velocity of violet light in the glass is given by

$v_{violet} = \frac{3(10^{8} )}{1.406}$

$v_{violet} =$ 2.13 × $10^{8} \ \frac{m}{s}$

This is the velocity of violet light in the glass.

8 0

2 years ago

What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is 10v/m?

Nitella [24]

Answer:

$3.33\cdot 10^{-8} T$

Explanation:

For an electromagnetic wave, the relationship between magnetic field amplitude and electric field amplitude is given by

$E=cB$

where

E is the amplitude of the electric field

c is the speed of light

B is the amplitude of the magnetic field

For the electromagnetic wave in this problem, we have

E = 10 V/m is the amplitude of the electric field

So if we solve the formula for B, we find the amplitude of the magnetic field:

$B=\frac{E}{c}=\frac{10 V/m}{3\cdot 10^8 m/s}=3.33\cdot 10^{-8} T$

4 0

3 years ago

The Cosmological Argument for the existence of God intellectually follows cause-and-effect back in time -- in effect asking "Whe

ankoles [38]

Answer:

First uncaused cause

Explanation:

Aristotle states that an infinite regression in the principle of causality is not possible. If the regression were infinite, then there would never be a first cause (mover), since this would need another mover to start its motion. Therefore, according to Aristotle, there must be an unmoved mover that moves other things, but is not itself moved by any prior action,

3 0

2 years ago

Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

3 years ago