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sashaice [31]
3 years ago
7

A student pulls horizontally on a 12 kg box, which then moves horizontally with an acceleration of 0.2 m/s^2. If the student use

s a force of 15 N, what is the coefficient of kinetic friction of the floor?
Physics
1 answer:
polet [3.4K]3 years ago
7 0
The net force of the object is equal to the force applied minus the force of friction. 
                         Fnet = ma = F - Ff
                           12 kg x 0.2 m/s² = 15 N - Ff
The value of Ff is 12.6 N. This force is equal to the product of the normal force which is equal to the weight in horizontal surface and the coefficient of friction.
                             Ff = 12.6 N = k(12 kg)(9.81 m/s²)
The value of k is equal to 0.107. 
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the distance between any two bodies is 10 M and the gravitational force between them is 3.2×10-⁹m. if the mass of one object is
cluponka [151]

Answer:

0.8 x 10^-9 kg

Explanation:

Given,

Distance ( R ) = 10 m

Force ( F ) = 3.2 x 10^-9 N

Mass ( m1 ) = 40 kg

To find : Mass ( m2 ) = ?

Formula : -

F = m1.m2 / R^2

m2 = FR^2 / m1

= 3.2 x 10^-9 x 10 / 40

= 3.2 x 10^-9 / 4

= ( 3.2 / 4 ) x 10^-9

m2 = 0.8 x 10^-9 kg

3 0
3 years ago
A railroad car is pulled through the distance of 960 m by a train that did 578 kJ of work during this pull.
Wittaler [7]

Answer:

<h2>602.08 N</h2>

Explanation:

The force supplied by the train can be found by using the formula

f =   \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{578000}{960}  \\  = 602.083333...

We have the final answer as

<h3>602.08 N</h3>

Hope this helps you

7 0
3 years ago
A football punker attempts to kick the football so that it lands on the ground 67.0 m from where it is kicked and stays in the a
Flauer [41]

To solve this problem we will apply the linear motion kinematic equations. We will find the two components of velocity and finally by geometric and vector relations we will find both the angle and the magnitude of the vector. In the case of horizontal speed we have to

v_x = \frac{x}{t}

v_x = \frac{67}{4.5}

v_x = 14.89m/s

The vertical component of velocity is

-h = v_y t -\frac{1}{2} gt^2

Here,

h = Height

g = Gravitational acceleration

t = Time

v_y = Vertical component of velocity

-1.23 = v_y(4.5)-\frac{1}{2} (9.8)(4.5)^2

-1.23= 4.5v_y - 99.225

v_y = 21.77m/s

The direction of the velocity will be given by the tangent of the components, then

tan\theta = \frac{v_y}{v_x}

\theta = tan^{-1} (\frac{21.77}{14.89})

\theta = 55.59\°

The magnitude is given vectorially as,

|V| = \sqrt{v_x^2+v_y^2}

|V| = \sqrt{14.89^2 +21.77^2}

|V| = 26.37m/s

Therefore the angle is 55.59° and the velocity is 26.37m/s

6 0
3 years ago
A ball is thrown vertically upward with an initial velocity
ivolga24 [154]

Answer:

D

Explanation:

First we define our variables

V0=29.4

a=-9.8

V=0

We have to find the maximum displacement , which I will define as X

We use formula v^2=v0^2+2aX

All we do is substitute our values

0=29.4^2-19.6X

29.4^2=19.6X

X=29.4^2/19.6=44.1

5 0
2 years ago
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

6 0
3 years ago
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