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professor190 [17]
3 years ago
10

A 4.4 kg object is being pushed along a surface, causing it to accelerate at a rate of 1.5 m/s2 . Th e coeffi cient of kinetic f

riction is 0.25. What is the magnitude of the horizontal force being applied to push the object?

Physics
1 answer:
emmasim [6.3K]3 years ago
8 0
Friction = 0.25 * m * g = 10.8
Push = m * a + Friction = 4.4 * 1.5 + 10.8

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If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
a person was using a spanner to loosen a tight nut.Their mass was 50kg the spanner was 0.5m long i) what is the persons weight?
sineoko [7]

1) Weight of the person: 490 N

2) Maximum torque: 245 Nm

Explanation:

1)

The weight of a body is equal to the gravitational force exerted on the body; it is given by the equation

F=mg

where

m is the mass of the body

g is the acceleration due to gravity

For the person in this problem,

m = 50 kg is the mass

g=9.8 m/s^2 is the acceleration due to gravity on Earth's surface

Therefore, the weight of the person is

F=(50)(9.8)=490 N

2)

The turning force (also called torque) exerted by a force rotating an object is given by

\tau = Fd sin \theta

where

F is the magnitude of the force

d is the length of the arm (the distance between the force and the pivotal point)

\theta is the angle between the direction of the force and the arm

For the spanner in this problem,

F = 490 N is the force applied (the weight of the person)

d = 0.5 m is the arm (the length of the spanner)

The maximum torque is obtained when \theta=90^{\circ}, therefore it is:

\tau=(490)(0.5)(sin 90^{\circ})=245 N\cdot m

Learn more about weight:

brainly.com/question/8459017

brainly.com/question/11292757

brainly.com/question/12978926

About torque:

brainly.com/question/5352966

#LearnwithBrainly

5 0
3 years ago
Rainwater draining from a neighborhood street initially travels at 4 ft/s through a pipe with a cross-sectional area of 15.7 ft2
Fudgin [204]

Answer:

The  velocity  is v_2  =  0.96 \ ft/s

Explanation:

From the question we are told that

   The initial speed is  v_1  =  4 \ ft/s

   The  cross -sectional area of the first pipe is  A_1  =  15.7 \ ft

   The  cross -sectional area of the second pipe is A_2 =  65.4 \  ft

Generally from continuity equation we have that

     A_1 * v_1 =  A_2  * v_2

So  

     v_2  =  \frac{A_1 *  v_1  }{A_2 }

=>   v_2  =  \frac{15.7  *  4  }{65.4 }

=>   v_2  =  0.96 \ ft/s

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Opposite chargers and like charges
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Opposite charges attract
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