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Anni [7]
3 years ago
15

Two persons are applying forces on two opposite sides of a moving cart. The cart still moves with the same speed in the same dir

ection. What do you infer about the magnitudes and direction of the forces applied?
Physics
1 answer:
masya89 [10]3 years ago
3 0
They are pushing on opposite sides against each other. The cart is moving forwards
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When Earth pulls on an object, that object also pulls on Earth. The values of these two forces are . This phenomenon can be expl
algol13
The values of these two forces are equal. Your weight on Earth is equal to the Earth's weight on you. When you and the Earth fall toward each other, your acceleration is greater than the Earth's acceleration, because your mass is less than the Earth's mass.
5 0
4 years ago
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A Tesla Roadster car accelerates from rest at a rate of 7.1m/s for a time of 3.9s
liberstina [14]

The distance covered is 54.0 m

Explanation:

Since the motion of the car is a uniformly accelerated motion, we can use the following suvat equation:

s=ut+\frac{1}{2}at^2

where

u is the initial velocity

t is the time

a is the acceleration

s is the distance covered

For the car in this problem, we have

u = 0 (it starts from rest)

a=7.1 m/s^2 is the acceleration

t = 3.9 s is the time

Substituting, we find s:

s=0+\frac{1}{2}(7.1)(3.9)^2=54.0 m

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5 0
4 years ago
Liam está empujando una caja pesada por una rampa a una velocidad constante. Hay fricción entre la rampa y la caja. La caja
Citrus2011 [14]

Answer:

Liam played danganrampa

Explanation:

5 0
3 years ago
A sanding disk with rotational inertia 2.0 x 10-3 kg·m2 is attached to an electric drill whose motor delivers a torque of magnit
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Answer

Given,

Rotational inertia = 2.0 x 10-3 kg·m²

Torque = 11 N.m

time, t = 19 ms

a) Angular momentum

  \tau = \dfrac{\Delta L}{\Delta t}

 L is angular momentum

  \Delta L = \tau \Delta t

  \Delta L = 11\times 19 \times 10^{-3}

  \Delta L = 0.209\ Kg m^2/s

b) Angular velocity

  We know.

     L = I \omega

     \omega = \dfrac{L}{I}

     \omega = \dfrac{0.209}{2\times 10^{-3}}

     \omega = 104.5\ rad/s

4 0
3 years ago
12. Taking the age of Earth to be about 4×10^9 years and assuming its orbital radius of 1.5 ×10^11 m has not changed and is circ
lana66690 [7]

The approximate total distance Earth has traveled since its birth is 3.8×10²¹m.

The Earth moves in a circular orbit of radius <em>r</em> around the Sun and it takes 1 year to go once around the Sun. Thus in a year it travels a distance equal to <em>2πr, </em>which is the circumference of the Earth's orbit.

Therefore in <em>n</em> years that have elapsed, it would travel a distance <em>d</em> given by,

d=2\pi rn

Substitute 3.14 for <em>π, </em>1.5×10¹¹m for <em>r</em> and 4×10⁹ for n.

d=2\pi rn\\= 2(3.14)(1.5*10^1^1m)(4*10^9)\\ =3.768*10^2^1m=3.8*10^2^1m

Thus, the approximate total distance Earth has traveled since its birth is 3.8×10²¹m.

7 0
3 years ago
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