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andrew-mc [135]
4 years ago
6

1. Accelerating objects are objects that are changing their velocity. Name the three controls on anautomobile that cause it to a

ccelerate.
Physics
1 answer:
Kay [80]4 years ago
3 0
Velocity is speed and its direction. Acceleration is any change of speed or direction. The gas pedal, the brake pedal, and the steering wheel can all cause that change.
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A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this
Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

p =  \frac{fd}{t}  \\ where \: \\p  = power \\  f = force \\ d = distance \\ and \: t = time \\  \\ p =  \frac{3800 \times 50}{10}  \\ p =  \frac{190000}{10}  \\ p = 19000w

7 0
4 years ago
if a man has a mass of 83 kilograms on earth, what will the force of gravity on his body be on the moon
AnnZ [28]

If a man has a mass of 83 kilograms on Earth, the force of gravity on his body be on the moon 135.6N. force =mass*acc , 83 * 9.8/6= 813.4/6 = 135.6N

3 0
3 years ago
Read 2 more answers
The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c
eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

X_f = X_i +\frac{1}{2}(V_i+V_f)t

Where,

X = Desplazamiento

V = Velocity

t = Time

In this case there is no initial displacement or initial velocity, therefore

X_f = \frac{1}{2} (V_i+V_f)t

Clearing for time,

t = \frac{2X_f}{(V_i+V_f)}

t = \frac{2*1.2}{24+0}

t = 0.1s

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

a= \frac{\Delta V}{t}

Then

F = m\frac{\Delta V}{t}

F = m\frac{V_f-V_i}{t}

F = m\frac{-V_i}{t}

F = \frac{(1600kg)(-24m/s)}{(0.1s)}

F = -384000N

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

V_f^2=V_i^2-2ax

0 = (24m/s)^2-2*a(1.2m)

a = \frac{(24m/s)^2}{1.2m}

a=480m/s^2

The gravity is equal to 0.8, then the acceleration is

a = 480*\frac{g}{9.8}

a = 53.3g

3 0
3 years ago
8. How did the measured angular magnification of the telescope compare with the theoretical prediction?
Genrish500 [490]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The theoretical angular magnification lies within the angular magnification range

Explanation:

From the question we are told that

   The  focal length of  B  is  f_{objective } =  43.0 \ cm

    The focal length of  A  is   f_{eye} =  10.4 \  cm

The  theoretical angular  magnification is mathematically represented as

           m = \frac{f_{objective }}{f_{eye}}  =  \frac{43.0}{10.4}

            m = \frac{f_{objective }}{f_{eye}}  =  4.175

Form the question the measured angular magnification ranges from 4 -5

So from the value calculated and the value given we can deduce that the theoretical angular  magnification lies within the angular magnification range

3 0
4 years ago
Weight of 1 kg becomes 1/6 on moon. If radius of moon is 1.76×10^6 m, then the mass of moon will be​
igomit [66]

Answer:

7.65 x 10^22kg

sorry if im wrong!

5 0
2 years ago
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