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irga5000 [103]
3 years ago
5

You throw a ball downward from a window at a speed of 2.5 m/s. how fast will it be moving when it hits the sidewalk 2.1 m below?

Physics
1 answer:
Otrada [13]3 years ago
4 0
List out all the variables that you do know; 

acceleration=-9.8 ms⁻¹ (this remains constant on Earth)
Final velocity=?
Displacement (s)= -2.1 m 
Initial Velocity(u)=2.5 ms⁻¹

v²=u²+2as 
v²=(2.5)²+2(-9.8)(-2.1)
v²=47.41 
v=√47.41 
v=6.88549 ≈ 6.9 ms⁻¹ 

Hope I helped :) 
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Answer:

V = IR

R = V/I

I = V/R

Explanation:

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2 years ago
A loaf of bread has a mass of 2000g and a volume of 500 cm3. what is the density of the bread
Yuki888 [10]

Answer:

4g/cm^3

Explanation:

density = mass / volume

            = 2000 / 500

            = 4g/cm^3

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2 years ago
Rachel has been reading her physics book. She takes her weighing scales into an elevator and stands on them. If her normal weigh
GrogVix [38]

Answer:

345 N

Explanation:

Given:

Normal weight of Rachel (mg) = 690 N

Case 1: Upward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₁ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=N_1-mg=N_1-690

Now, from Newton's second law:

F_{net}=ma\\\\N_1-690=m\times 0.25g\\\\N_1-690=0.25\times (mg)\\\\N_1-690=0.25\times 690\\\\N_1=690+172.5=862.5\ N------(1)

Case 2: Downward motion of elevator

Given:

Acceleration of elevator (a) = 0.25 g

The scale reading is given by the normal force acting on Rachel. Let N₂ be the normal force.

So, the net force acting on Rachel is given as:

F_{net}=mg-N_2=690-N_2

Now, from Newton's second law:

F_{net}=ma\\\\690-N_2=m\times 0.25g\\\\690-N_2=0.25\times (mg)\\\\690-N_2=0.25\times 690\\\\N_2=690-172.5=517.5\ N------(2)

Now, the difference in the scale reading is obtained by subtracting equation (2) from equation (1). This gives,

Difference=N_1-N_2=862.5-517.5=345\ N

Therefore, the difference between the up and down scale readings is 345 N.

4 0
2 years ago
A horizontal disk with a radius of 10 cm rotates about a vertical axis through its center. The disk starts from rest at t = 0 an
Tom [10]

Answer:

0.69s

Explanation:

10 cm = 0.1 m

Let t be the time that radial and tangential components of the linear acceleration of a point on the rim be equal in magnitude. At that time we have the angular velocity would be

\omega = \alpha t = 2.1 t

And so the radial acceleration is

a_r = \omega^2 r = (2.1t)^2 r = 2.1^2 t^2 * 0.1= 0.441 t^2 m/s^2

The tangential acceleration is always the same since angular acceleration is constant:

a_t = \alpha * r = 2.1 * 0.1 = 0.21 m/s^2

For these 2 quantities to be the same

a_r = a_t

0.441 t^2 = 0.21

t^2 = 0.21/0.441 = 0.4762

t = \sqrt{0.4762} = 0.69 s

6 0
3 years ago
What acceleration occurs if the friction force in #4 is now 4N?
anastassius [24]

Answer:

you a b1tc h

Explanation:

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2 years ago
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