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3 years ago

A book is resting on the table. Forces are exerted _____.

Physics

2 answers:

tino4ka555 [31]3 years ago

4 0

for the last one, its 2 m/s2

scoray [572]3 years ago

3 0

Forces are exerted I believe : all of the above
The action force might be Tyler throwing the ball
I don't know the last one

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A 2.0-kg block slides on a rough horizontal surface. A force (magnitude P = 4.0 N) acting parallel to the surface is applied to

irinina [24]

When the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

<h3>Frictional force between the block and the horizontal surface</h3>

The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

5 - 1.6 = 2a

3.4 = 2a

a = 3.4/2

a = 1.7 m/s²

Thus, when the applied force increases to 5 N, the magnitude of the block's acceleration is 1.7 m/s².

Learn more about frictional force here: brainly.com/question/4618599

8 0

2 years ago

Which letter represents the location of the<br> resister in this diagram?<br> A<br> B<br> C<br> D

ololo11 [35]

Answer:

The letter B is the letter that represents the location of the resister in the diagram.

Explanation:

I hope it helped!Please mark brainliest and have a wonderful night! and day!!!!

5 0

3 years ago

Read 2 more answers

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

4 0

3 years ago

An inductor is connected to the terminals of a battery that has an emf of 16.0 V and negligible internal resistance. The current

balu736 [363]

Answer:

(a) The resistance R of the inductor is 2480.62 Ω

(b) The inductance L of the inductor is 1.67 H

Explanation:

Given;

emf of the battery, V = 16.0 V

current at 0.940 ms = 4.86 mA

after a long time, the current becomes 6.45 mA = maximum current

Part (a) The resistance R of the inductor

$R = \frac{V}{I_{max}} = \frac{16}{6.45*10^{-3}} = 2480.62 \ ohms$

Part (b) the inductance L of the inductor

$\frac{Rt}{L} = -ln(1-\frac{I}I_{max}})\\\\L = \frac{Rt}{-ln(1-\frac{I}I_{max})}}$

where;

L is the inductance

R is the resistance of the inductor

t is time

$L = \frac{Rt}{-ln(1-\frac{I}I_{max})}} = \frac{2480.62*0.94*10^{-3}}{-ln(1-\frac{4.86}{6.45})} \\\\L =\frac{2.3318}{1.4004} = 1.67 \ H$

Therefore, the inductance is 1.67 H

5 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago