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stepan [7]
3 years ago
14

A student is getting ready to go on a bike ride, but her bike back tire is flat! She connects a pump to the valve and adds air t

o the tire until it is fully inflated and ready to ride. After the addition of air to the tire, the air pressure inside the tire is now __________ the air pressure outside the tire. Fill in the blank by choosing one answer below.\
Physics
1 answer:
Mariana [72]3 years ago
4 0
Wheres the answer below?
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PLEASE HELP In a bag are 7 red, 9 blue, 2 yellow and 4 green marbles. If you draw out a marble at random, what the probability t
Free_Kalibri [48]

Answer:

16/22

Explanation:

you add red and blue together

6 0
3 years ago
Future space stations could create an artificial gravity by rotating. Consider a cylindrical space station that rotates with a p
aleksandrvk [35]

Answer:

P = 2 pi R / v    period of space station

F / m = v^2 / R    centripetal force per unit of mass

So F / m = 4 pi^2 R^2 / (P^2 * R) = 4 pi^2 R / P^2

Also, F / m = 9.8 m/s^2   earth's gravitational attraction

So 9.8 = 4 pi^2 R / P^2    or    R = 9.8 P^2 / 4 * pi^2) = 195 m

Or D = 2 R = 390 m the diameter required

8 0
3 years ago
Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi
Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

   P₁ = m v

   P₁ = 0.0575 x 26.7

   P₁ = 1.535 kg.m/s

b) momentum of ball after collision

   P₂ = m v'

   P₂ = 0.0575 x (-29.5)

   P₂ = -1.696 kg.m/s

c) change in momentum

    Δ P = P₂ - P₁

    Δ P = -1.696 -1.535

    Δ P = -3.231 kg.m/s

d) using conservation of momentum

  initial speed of racket = 0 m/s

  M u + m v = Mu' + m v

  M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

  0.331 u' = 3.232

     u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

5 0
4 years ago
1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force
Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

4 0
4 years ago
Kali left school and traveled toward her friend's house at an average speed of 40 km/h. Matt left one hour later and traveled in
zlopas [31]

Answer:

t = 5 hr

Explanation:

Let kali moves toward east with velocity= V₁= 40 km/ h

Mat moves toward west with velocity = V₂= 50 km/hr

As Klai left one hour earlier = t₁= 1 hr

distance traveled in 1st hour = s₁ = v * t = 40 * 1 = 40 km

Remaining distance = 400 - 40 = 360 km

As they move in the opposite directions:

Relative speed= 40 + 50 = 90 km/ h

s = v * t

⇒ t = s / v

⇒ t₂ = 360 / 90

⇒ t₂ = 4 hr

Total time = t = t₁ + t₂

t = 1 hr + 4 hr

t = 5 hr

5 0
3 years ago
Read 2 more answers
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