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Elza [17]
3 years ago
7

A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th

e sled is doubled, what is the acceleration of the sled?
Physics
1 answer:
mihalych1998 [28]3 years ago
4 0
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
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A plane traveled for 3 hours at a velocity of 1200 km/hr North. What is the distance traveled?
Daniel [21]

Answer:

the answer is 3600 kilometers

4 0
3 years ago
A biker pedals hard to ride his bike to the top of a 44 m hill. He decides to let his bike coast down the hill, and is having so
Dafna11 [192]

Answer:

The bikers speed at the top of other hill is <u>25.82 m/s.</u>

Explanation:

Considering the biker is riding on a frictionless surface.

∴ There is no non-conservative or external force acting on the biker.

Hence we can conserve the energy of biker and bike as a system.

Let,

h_{1} = 44m

h_{2} = 10m

Since the biker starts from rest , his initial speed v_{1} = 0 m/s

Let final speed of the bike at the top of other hill be v_{2}.

∴ Initial Energy (at the top of 44m hill) = mgh_{1}

  Final Energy  (at the top of 10m hill) =  mgh_{2} + \frac{1}{2}mv_{2} ^{2}.

Conserving both the energies , we get

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2} ^{2}

∴ v_{2} = \sqrt{2g(h_{1}-h_{2} )}

Substituting the values for g , h_{1} , h_{2} , we get

v_{2} = 25.82 m/s

6 0
3 years ago
A day on a distant planet observed orbiting a nearby star is 21.5 hr. Also, a year on the planet lasts 69.3 Earth days. In other
serg [7]

Answer:

Part A

The angular speed of rotation of the plane is 8.11781 × 10⁻⁵ rad/s

Part B

The angular speed of orbit of the planet is 1.04938 × 10⁻⁶ rad/s

Explanation:

The parameters of the planet are;

The duration of a day on the distant planet = 21.5 hr.

The duration of a year on the distant planet = 69.3 Earth days

Part A

The duration of a day = The time to make one complete revolution of 2·π radians

∴ The average angular speed about its axis, \omega_{rotation} = Angle turned/Time

∴ \omega_{rotation}  = 2·π/(21.5 × 60 × 60) s ≈ 8.11781 × 10⁻⁵ rad/s

The average angular speed of the planet about its own axis, \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

The angular speed of rotation of the plane \omega_{rotation}  = 8.11781 × 10⁻⁵ rad/s

Part B

The time it takes the planet to revolve round the neighboring star once = 69.3 Earth days

Therefore, the average angular speed of the planet around its neighboring star, \omega _{Star}, is given as follows;

\omega _{Orbit}  = 2·π/((69.3 × 24 × 60 × 60) s) = 1.04938 × 10⁻⁶ rad/s

The average angular speed of orbit, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s

The angular speed of orbit of the planet, \omega _{Orbit} = 1.04938 × 10⁻⁶ rad/s.

3 0
2 years ago
a 20 A fuse is connected in series with a circuit containing a 240 V source. What is the minimum resistance required to prevent
kipiarov [429]

Answer:

12 ohms

Explanation:

you just divide 240 and 20

240 V/ 20 A = 12 ohms

7 0
3 years ago
When an acid reacts with water, the hydroxide ion is formed.<br><br> True<br><br> False
Marysya12 [62]
When acid reacts with water, the hydroxide ion is formed I think it is true.
4 0
3 years ago
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