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natulia [17]
4 years ago
5

You can also enter units that are combinations of other units. Keep in mind that you have to indicate the multiplication of unit

s explicitly either with a multiplication dot or a dash. You can add a multiplication dot by selecting the dot There is a dot button. from the Templates There is a templates button. menu. The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). Of an object’s mass is m=10. kg, what is its weight
Physics
1 answer:
juin [17]4 years ago
8 0

Answer:

98 N

Explanation:

Given that The weight of an object is the product of its mass, m, and the acceleration of gravity, g (where g=9.8 m/s2). Of an object’s mass is m=10. kg,

The parameters to be considered are

Mass m = 10 kg

Acceleration due to gravity = 9.8m/s^2

Weight W = mg

substitute mass m and acceleration due to gravity g into the formula above.

Weight = 10 × 9.8

Weight = 98 N

Therefore, the weight of the body or object is 98 N

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3 years ago
g While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll
irga5000 [103]

Answer:

The minimum coefficient of static friction required, µ = 0.10

<em>Note. The question is incomplete. The complete question is given below:</em>

<em>While hauling a log in the back of a flatbed truck, a driver is pulled over by the state police. Although the log cannot roll sideways, the police claim that the log could have slid out the back of the truck when accelerating from rest. The driver claims that the truck could not possibly accelerate at the level needed to achieve such an effect. Regardless, the police write a ticket anyway and now the driver court date is approaching.</em>

<em>The log has a mass of m = 929 kg; the truck has a mass of M = 8850 kg. According to the truck manufacturer, the truck can accelerate from 0 to 55 mph in 23.0 seconds, but this does not account for the additional mass of the log. Calculate the minimum coefficient of static friction μs needed to keep the log in the back of the truck.</em>

Explanation:

First, velocity in mph is converted to m/s

1 mph = 0.447 m/s

55 mph ≈ 24.6 m/s

The acceleration of the empty truck is a = v/t = 24.6 / 23 = 1.07 m/s²

Force that can be generated by the truck, F = ma

F = 8850kg * 1.07 m/s² = 9469.5 N

However, with the added mass of the log on it, the acceleration of the truck will become;

a = F / m = 9469.5 N / (8550+929)kg = 0.97 m/s²

Frictional force between the log and the truck = 0.97 m/s² * 929 kg = 901.13 N

Normal reaction on the truck due to the weight of the log, R = mg

R = 929 kg * 9.8m/s² = 9104.2 N

Coefficient of static friction, µ = F/R

µ = 901.13/9104.2

µ = 0.098 ≈ 0.10

Therefore, the minimum static friction required is µ = 0.10

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3 years ago
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A train going 14m/s moves 250 m while accelerating to a stop. What is the train’s deceleration?
Elanso [62]

Answer:

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3 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

mg sin \theta - \mu N = ma (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, \theta=39^{\circ} being the angle of the incline

\mu N is the frictional force, with \mu = 0.6 being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

N-mg cos \theta =0

where mg cos \theta is the component of the weight in the direction perpendicular to the slope. Solving for N,

N=mg cos \theta

Substituting into (1), solving for a, we find the acceleration:

a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2

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