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timama [110]
3 years ago
9

Describe what happens to the speed of a bicycle as it goes uphill and downhill.

Physics
1 answer:
Elenna [48]3 years ago
7 0
Describe what happens to the speed of a bicycle as it goes downhill
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Find the force in newtons that does 0.0284 kilojoules of work in moving a book a distance of 4.00 meters
gulaghasi [49]

Work = (force) x (distance

28.4 joules = (force) x (4 meters)

Divide each side by (4 meters) :

Force = (28.4 joules) / (4 meters)

Force = 7.1 Newtons

5 0
4 years ago
What's the kinetic energy of an object that has a mass of 30 kilograms and moves with a velocity of 20m/s?
icang [17]

Ek = 6KJ.

In physics, the kinetic energy of a body or object is the one that owns due to its movement and is given by the equation E_{k} = \frac{1}{2} mv^{2}, where m is the mass of the object in kilograms and v is the velocity in m/s.

An object that it has a mass of 30 kilograms and moves with a velocity of 20m/s, its kinetic energy is given by:

E_{k} = \frac{1}{2} (30kg)(20m/s)^{2}=6000J=6KJ

4 0
3 years ago
The magnitude of the velocity of a projectile when it is at its maximum height above ground level is 14 m/s. (a) What is the mag
Alecsey [184]

When the projectile is at its maximum height above ground, it's at the point
of changing from rising to falling.  At that exact point, its vertical speed is zero,
so the 14 m/s must be all horizontal velocity.  That's not going to change.

Since we need to consider changes in vertical speed now, we need to make
some assumption about where this is all happening, so that we know the
acceleration of gravity.  I'll assume that it's all happening on or near the Earth,
and the acceleration of gravity is 9.8 m/s².

I'm also going to neglect air resistance.

a). 1.2 sec before it reaches its maximum height, the projectile is rising
at a vertical speed of (1.2 x 9.8) = 11.76 m/s. 
The magnitude of its velocity is

the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° above horizontal.

b).  1.2 sec after it reaches its maximum height, the projectile is falling
at a vertical speed of (1.2 x 9.8) = 11.76 m/s. 
The magnitude of its velocity is

the square root of (14² + 11.76²) = 18.28 m/s, directed about 40° below horizontal.

===========================

In 1.2 second before or after zero vertical speed, an object in free fall moves

         (1/2) (g) (t²) = (4.9) (1.2²) = 7.06 meters .

c). & d).
1.2 seconds before it reaches maximum height, the projectile is located at

           x = -14 m
           y = -7.06 m

e). & f).
1.2 seconds after it reaches maximum height, the projectile is located at

           x = +14 m
           y = -7.06 m .


I hope you recognize that 6 answers, plus a little bit of explanation,
all for 5 points, ain't too shabby.  You made out well.

5 0
3 years ago
Commercially-available hybrid vehicles, such as the Toyota Prius, use electrical batteries to store energy for later use. Howeve
Bad White [126]

(A) 4.2\cdot 10^5 J

The energy stored by the system is given by

E=Pt

where

P is the power provided

t is the time elapsed

In this case, we have

P = 60 kW = 60,000 W is the power

t = 7 is the time

Therefore, the energy stored by the system is

E=(60,000 W)(7 s)=4.2\cdot 10^5 J

(B) 4830 rad/s

The rotational energy of the wheel is given by

E=\frac{1}{2}I \omega^2 (1)

where

I is the moment of inertia

\omega is the angular velocity

The moment of inertia of the wheel is

I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2

where M is the mass and R the radius of the wheel.

We also know that the energy provided is

E=4.2\cdot 10^5 J

So we can rearrange eq.(1) to find the angular velocity:

\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s

(C) 2.8\cdot 10^6 m/s^2

The centripetal acceleration of a point on the edge is given by

a=\omega^2 R

where

\omega=4830 rad/s is the angular velocity

R = 0.12 m is the radius of the wheel

Substituting, we find

a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2

7 0
4 years ago
According to the concept of length contraction, what happens to the length of an object as it approaches the speed of light and
lord [1]
When an object moves its length contracts in the direction of motion. The faster it moves the shorter it gets in the direction of motion.
The object in this question moves and then stops moving. So it's length first contracts and then expands to its original length when the motion stops.
The speed doesn't have to be anywhere near the speed of light. When the object moves its length contracts no matter how fast or slow it's moving.
8 0
4 years ago
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