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professor190 [17]
3 years ago
15

You need to get to class, which is 282 meters away. You can only walk in the hallway at about 2.0 m/s (if you go any faster, you

'll be caught or running).
How much time will it take to get to your class?
Physics
1 answer:
worty [1.4K]3 years ago
6 0

Answer:

Time, t = 141 seconds

Explanation:

Given that,

Distance between the class and the person, d = 282 m

Speed of the person, v = 2 m/s

We need to find the time taken by the person to get to your class. Let the time taken is t. The speed of the person is given by :

v=\dfrac{d}{t}

t=\dfrac{d}{v}

t=\dfrac{282\ m}{2\ m/s}

t = 141 seconds

So, the time taken by the person to get to your class is 141 seconds. Hence, this is the required solution.

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iris [78.8K]

me ajudem por favor pra agora de noite

5 0
3 years ago
A 235 kg crate is pulled across a horizontal surface with a force of 760 N applied at an
AnnyKZ [126]

Answer:

658.16N

Explanation:

Step one:

given data

mass m= 235kg

Force F= 760N

angle= 30 degrees

Required

The horizontal component of the force

Step two:

The horizontal component of the force

Fh= 760cos∅

Fh=760cos30

Fh=760*0.8660

Fh=658.16N

3 0
3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f
Hitman42 [59]
(a) The angular acceleration of the wheel is given by
\alpha =  \frac{\omega_f - \omega_i }{t}
where \omega_i and \omega_f are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
\alpha =  \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
\theta (t) = \omega_i t +  \frac{1}{2} \alpha t^2
where \omega_i = 0 is the initial angular speed. So, the angle covered after a time t=3.07 s is
\theta=  \frac{1}{2}  \alpha t^2 =  \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad
6 0
3 years ago
A 1400-kg car is traveling with a speed of 17.7 m/s. What is the magnitude of the horizontal net force that is required to bring
AleksandrR [38]

Answer:

The magnitude of the horizontal net force is 13244 N.

Explanation:

Given that,

Mass of car = 1400 kg

Speed = 17.7 m/s

Distance = 33.1 m

We need to calculate the acceleration

Using equation of motion

v^2-u^2=2as

Where, u = initial velocity

v = final velocity

s = distance

Put the value in the equation

0-(17.7)^2=2a\times 33.1

a=\dfrac{-(17.7)^2}{33.1}

a=-9.46\ m/s^2

Negative sign shows the deceleration.

We need to calculate the net force

Using newton's formula

F = ma

F =1400\times(-9.46)

F=-13244\ N

Negative sign shows the force is opposite the direction of the motion.

The magnitude of the force is

|F| =13244\ N

Hence,  The magnitude of the horizontal net force is 13244 N.

5 0
3 years ago
The force that attracts earth to an object is equal to and opposite the force that earth exerts on the object. Explain why earth
alexandr402 [8]

Answer:

Because of heavy mass

Explanation:

When force acts on a body it tends to accelerate the body. The acceleration produced in the body depends on two things:

1). Magnitude of force

2). Mass of the body

F= ma

⇒ a = F/m  

As the force exerted on earth and another object are the equal in magnitude but opposite in direction. This forces will accelerate the object toward the earth but can't accelerate the earth as earth has very high mass.

a = F/m

This force tends to accelerate the earth but but due to earth's inertia the earth does not accelerate.

7 0
3 years ago
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