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3 years ago

While driving down the road, an unfortunate bug strikes the windshield of a bus

Physics

1 answer:

ivolga24 [154]3 years ago

3 0

Answer: Both have the same amount of force each exerts.

Explanation: Newton's third law of motion known as the action-reaction states that the amount of force exerted by an object is equal to the force exerted against the other but only differ in the direction of the force which is opposite. Fa = -Fb as indicated. The bug pushes the windshield it is the action, as well as the windshield pushes the bug the reaction. So both exert same amount of force.

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List the number of atoms for each element in this compound: $CaCO_{3} $

borishaifa [10]

Number of element atom for element,

Calcium (Ca) = 1

Carbon (C) = 1

Oxygen (O) = 3

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3 years ago

If a ray of light traveling in the liquid has an angle of incidence at the interface of 33.0 ∘, what angle does the refracted ra

aliina [53]

Answer:

29°

Explanation:

because the refracted ray angle is small than angle of incidence

8 0

3 years ago

Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

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3 years ago

A 2-f and a 1-f capacitor are connected in series and a voltage is applied across the combination. the 2-f capacitor has:_______

VARVARA [1.3K]

Half the potential difference of the the1-µF

A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.

The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.

Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.

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2 years ago

Read 2 more answers

A 50 kg student climbs 3m to the top of a set of stairs. Calculate the change in the student’s gravitational potential energy fr

erastova [34]

Gpe = mgh
gpe = 50*10*3
gpe = 1500 J

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3 years ago