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kondor19780726 [428]
3 years ago
12

How long would it take a person to come to a stop if they were going

Physics
1 answer:
hammer [34]3 years ago
8 0
33 mins hope it’s right
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Alkaline Earth Metals is that one
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What are some other examples of a force acting on an object without<br> touching it?
a_sh-v [17]
An example would be gravity
8 0
2 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car  = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

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

s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2

now, equating both the distances

23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2

t_2^2 = 15.143

    t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s

4 0
2 years ago
The upward force exerted on an object falling through air is _____.
stira [4]

(C) Air Resistance

<u>Explanation:</u>

When an object falls through air, air resistance acts on it in upward direction. When air resistance acts, acceleration during a fall will be less than g because air resistance affects the motion of the falling objects by slowing it down. Air resistance depends on two important factors - the speed of the object and its surface area. Increasing the surface area of an object decreases its speed.

6 0
3 years ago
A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the
Shalnov [3]

Answer:

x ’= 1,735 m,  measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

             

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

           x_{cm} = 1.2 -1

          x_ {cm} = 0.2 m

          Σ τ = 0

          w₁ 1.2 + mg 0.2 - W₂ x = 0

          x = \frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}

          x = \frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}

let's calculate

          x = \frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}2.9 1.2 + 4 0.2 / 8

           

          x = 0.535 m

measured from the pivot point

measured from the far left is

           x’= 1,2 + x

           x'=  1.2 + 0.535

           x ’= 1,735 m

8 0
2 years ago
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