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IRISSAK [1]
3 years ago
11

Calculate the force of friction for a 5kg aluminum block being pulled with constant velocity (uniform motion) across a horizonta

l steel plank. ...?
Physics
1 answer:
Stolb23 [73]3 years ago
4 0
The purpose of this lab is to determine whether the surface of an area would affect the coefficient of Friction. My classmates and I have learned a lot in this lab and that there could have been some errors in our lab because the strength of how a person pulls it might be a slight different than the normal force. I learned from this lab that the <span>surface area would have no effect on the coefficient of friction. </span>
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Which of the following is occurring while a satellite is in orbit around Earth? O It is continuously pulling away from Earth It
hoa [83]

AnswerIt is continuously falling towards the surface of the earth

Explanation:

since gravity from earth is the thing that keeps it constantly in orbit

4 0
3 years ago
A 2200 kg car starts from rest and speeds up to 12 m/s in 5.2 s. The net force acting on the car is?
zmey [24]
The acceleration is 12(m/s)/5.2s = 2.308m/s^2
F=ma, so
F=2200*2.308 = 5076.9N
6 0
3 years ago
How does changing the voltage in a circuit affect the current in the circuit
Tasya [4]
The formula for the voltage is shown below:
Voltage = Current x Resistance
V=I x R

When we change voltage:
I =V/R or R=V/I
Increasing the voltage in an electrical circuit will also increase the current value, as well as the resistance of the load,
 
7 0
3 years ago
Read 2 more answers
Physics Homework MathPhys homie if you see this pls help
cluponka [151]

Answer:

1. -8.20 m/s²

2. 73.4 m

3. 19.4 m

Explanation:

1. Apply Newton's second law to the car in the y direction.

∑F = ma

N − mg = 0

N = mg

Apply Newton's second law to the car in the x direction.

∑F = ma

-F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

Given μ = 0.837:

a = -(9.8 m/s²) (0.837)

a = -8.20 m/s²

2. Given:

v₀ = 34.7 m/s

v = 0 m/s

a = -8.20 m/s²

Find: Δx

v² = v₀² + 2aΔx

(0 m/s)² = (34.7 m/s)² + 2 (-8.20 m/s²) Δx

Δx = 73.4 m

3. Since your braking distance is the same as the car in front of you, the minimum safe following distance is the distance you travel during your reaction time.

d = v₀t

d = (34.7 m/s) (0.56 s)

d = 19.4 m

6 0
4 years ago
A ball is in free fall after being dropped. What willthe speed of the ball be after 2 seconds of free fall?
Mazyrski [523]

So, the speed of the ball after 2 seconds after free fall is <u>20 m/s</u>.

<h3>Introduction</h3>

Hi ! I'm Deva from Brainly Indonesia. In this material, we can call this event "Free Fall Motion". There are two conditions for free fall motion, namely falling (from top to bottom) and free (without initial velocity). Because the question only asks for the final velocity of the ball, in fact, we may use the formula for the relationship between acceleration and change in velocity and time. In general, this relationship can be expressed in the following equation :

\boxed{\sf{\bold{a = \frac{v_2 - v_1}{t}}}}

With the following conditions :

  • a = acceleration (m/s²)
  • \sf{v_2} = speed after some time (m/s)
  • \sf{v_1} = initial speed (m/s)
  • t = interval of time (s)

<h3>Problem Solving</h3>

We know that :

  • a = acceleration = 9,8 m/s² >> because the acceleration of a falling object is following the acceleration of gravity (g).
  • \sf{v_1} = initial speed = 0 m/s >> the keyword is free fall
  • t = interval of time = 2 s

What was asked :

  • \sf{v_2} = speed after some time = ... m/s

Step by step :

\sf{a = \frac{v_2 - v_1}{t}}

\sf{(a \times t) + v_1 = v_2}

\sf{(10 \times 2) + 0 = v_2}

\boxed{\sf{v_2 = 20 \: m/s}}

So, the speed of the ball after 2 seconds after free fall is 20 m/s.

8 0
2 years ago
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