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mario62 [17]
3 years ago
7

Need help with these 4 questions

Physics
1 answer:
dimaraw [331]3 years ago
4 0

Answer: 1. B

2.A

3.B

4. A

Explanation:

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If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric
poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0
3 years ago
A plasma wave moving through a plasma has a frequency of 109 Hz and a speed of 3.0 × 107 m/s. What is the wavelength of this wav
tigry1 [53]

Answer:0.27*10^{6} (m)

Explanation:

with a wave we know that  speed= wavelenght* frecuency   so

wavelenght= 3*10^{7} / 109  = 0.27*10^{6} (m)

8 0
3 years ago
A 16.5-kg crate starts at rest at the top of a 60.0° incline. The coefficients of friction are μs = 0.400 and μk = 0.300. The cr
irga5000 [103]

Answer:

t = 1.62 s

Explanation:

given,

mass of the block m₁ = 16.5 Kg

m₂ = 8 Kg

angle of inclination = 60°

μs = 0.400 and μk = 0.300

time to slide 2 m = ?

a) let a is the acceleration of the block m₁ downward.

Net force acting on m₂,

F₂ = T - m₂ g

m₂a = T - m₂ g

a = \dfrac{T}{m_2} - g.......(1)

net force acting on m₁

F₁ = m₁g sin(60°) - μ_k m₁g cos (60°) - T

m₁ a = m₁g sin(60°) - μ_k m₁g cos (60°) - T

a = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}.........(2)

from equations 1 and 2

\dfrac{T}{m_2} - g = g sin(60^0) - \mu_k g cos (60^0) - \dfrac{T}{m_1}

\dfrac{T}{m_2} +\dfrac{T}{m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T\dfrac{m_1+m_2}{m_2\times m_1} = g+ g sin(60^0) - \mu_k g cos (60^0)

 T = \dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{\dfrac{m_1+m_2}{m_2\times m_1}}

 T = {m_2\times m_1}\dfrac{g+ g sin(60^0) - \mu_k g cos (60^0)}{{m_1+m_2}}  

T = {16.5\times 8}\dfrac{9.8 + 9.8 sin(60^0) - 0.3\times 9.8 cos (60^0)}{{16.5+8}}

T = 90.61 N

from equation (1)

a = \dfrac{90.61}{8} - 9.8.......(1)

a = 1.52 m/s²

let t is the time taken

Apply,

d = ut + 0.5 a t²

2 = 0 + 0.5 x 1.52 x t²

t = \sqrt{2.63}

t = 1.62 s

5 0
3 years ago
Two charged objects, A and B, are exerting an electric force on each other. What will happen if the charge on A is increased?
Nina [5.8K]
The force is proportional to the product of the charges. So if either one increases, the force between them increases.
7 0
4 years ago
Read 2 more answers
If the x-component of velocity is 27m/s and the y-component of velocity is -23 m/s, what is the resultant vector?
Dafna11 [192]

Answer:

35.47 m/s

Explanation:

r =  \sqrt{( - 23)^{2}  + (27) ^{2} }

=35.57

5 0
4 years ago
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