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forsale [732]
3 years ago
15

I NEED THE ANSWER TO THIS SO I CAN GET MY PHONE BACK

Physics
1 answer:
Setler [38]3 years ago
4 0
The answer is C I not sure
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When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi
SVEN [57.7K]

when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

f = \frac{mv - mu}{t}

v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}

since the ball is going down, we take that as negative and ball going upwards as positive.

f = 189 N

the force exerted from the floor is 189 N

4 0
3 years ago
Select the correct answer.
Alenkasestr [34]
The answer would be c because it is talking about she wants to be in a good neighborhood
3 0
3 years ago
Read 2 more answers
A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th
Aleksandr [31]

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

7 0
3 years ago
A small box slides down a ramp on a friction with surface. If the total energy of the system is conserved, which computational m
spayn [35]

Answer:

WWWEEEEEEEEEEEEEEEEEE

Explanation:

6 0
3 years ago
Read 2 more answers
A 5-cm-high peg is placed in front of a concave mirror with a radius of curvature of 20 cm.
Andrei [34K]

Answer:

Explanation:

Using the magnification formula.

Magnification = Image distance(v)/object distance(u) = Image Height(H1)/Object Height(H2)

M = v/u = H1/H2

v/u = H1/H2...1

3) Given the radius of curvature of the concave lens R = 20cm

Focal length F = R/2

f = 20/2

f = 10cm

Object distance u = 5cm

Object height H2= 5cm

To get the image distance v, we will use the mirror formula

1/f = 1/u+1/v

1/v = 1/10-1/5

1/v = (1-2)/10

1/v =-1/10

v = -10cm

Using the magnification formula

(10)/5 = H1/5

10 = H1

H1 = 10cm

Image height of the peg is 10cm

4) If u = 15cm

1/v = 1/f-1/u

1/v = 1/10-1/15

1/v = 3-2/30

1/v = 1/30

v = 30cm

30/15 = H1/5

15H1 = 150

H1/= 10cm

5) if u = 20cm

1/v = 1/f-1/u

1/v = 1/10-1/20

1/v = 2-1/20

1/v = 1/20

v = 20cm

20/20 = H1/5

20H1 = 100

H1 = 5cm

6) If u = 30cm

1/v = 1/f-1/u

1/v = 1/10-1/30

1/v = 3-1/30

1/v = 2/30

v = 30/2 cm

v =>15cm

15/30 = Hi/5

30H1 = 75

H1 = 75/30

H1 = 2.5cm

4 0
3 years ago
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