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stepan [7]
2 years ago
14

A constant unbalanced force of friction acts on a 15.0-kilogram mass moving along a horizontal surface at 10.0 meters per second

. If the mass is brought to rest in 1.50 seconds, what is the magnitude of the force of friction
Physics
1 answer:
bearhunter [10]2 years ago
5 0

The mass slows to a rest from an initial speed of 10.0 m/s in a matter of 1.50 s. If we assume constant acceleration, then the mass has acceleration a such that

0 = 10.0 m/s + a (1.50 s)   ⇒   a ≈ -(10.0 m/s)/(1.50 s)

The net force acting on the mass as it slows down is

∑ F[horizontal] = -F[friction] = ma

where m = 15.0 kg, and we take the direction in which friction is acting to be negative. Then

-F[friction] = -(15.0 kg) (10.0 m/s)/(1.50 s)   ⇒   F[friction] = 100 N

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A bus driver heads south with a steady speed of
Nutka1998 [239]

ANS

{ A }

Magnitude

5400.097 \: ✓ \: m

Direction

25.52 \:  ✓ \: ° South  \: of \:  West

{ B }

Average speed in ( m/s )

23.43 \:  ✓  \: m/s

{ C }

Magnitude

14.06 \:  ✓  \: m/s

Direction

25.52 \:  ✓  \: ° South  \: of \:  West

5 0
3 years ago
A 25N force is acting on a body moving on a straight line with Initial momentum 20 kam's. Find the final momentum after 4 second
Nezavi [6.7K]

The final momentum of the body is equal to 120 Kg.m/s.

<h3>What is momentum?</h3>

Momentum can be described as the multiplication of the mass and velocity of an object. Momentum is a vector quantity as it carries magnitude and direction.

If m is an object's mass and v is its velocity then the object's momentum p is: {\displaystyle \mathbf {p} =m\mathbf {v} . The S.I. unit of measurement of momentum is kg⋅m/s, which is equivalent to the N.s.

Given the initial momentum of the body = Pi = 20 Kg.m/s

The force acting on the body, Pf = 25 N

The time, Δt = 4-0 = 4s

The Force is equal to the change in momentum: F ×Δt = ΔP

25 × 4 = P - 20

100 = P - 20

P = 100 + 20 = 120  Kg.m/s

Therefore, the final momentum of a body is 120 Kg.m/s.

Learn more about momentum, here:

brainly.com/question/4956182

#SPJ1

5 0
1 year ago
The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pre
IrinaVladis [17]

Answer:

A) T1 = 269.63 K

T2 = 192.59 K

B) W = -320 KJ

Explanation:

We are given;

Initial volume: V1 = 7 m³

Final Volume; V2 = 5 m³

Constant Pressure; P = 160 KPa

Mass; m = 2 kg

To find the initial and final temperatures, we will use the ideal gas formula;

T = PV/mR

Where R is gas constant of helium = R = 2.0769 kPa.m/kg

Thus;

Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K

Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K

B) world one is given by the formula;

W = P(V2 - V1)

W = 160(5 - 7)

W = -320 KJ

6 0
3 years ago
Step 2: Apply NEwton's second law Apply ∑Fy = may , what should ay be equal to, since the block doesn't move in the y direction
andrey2020 [161]

Answer:

∑Fy = 0, because there is no movement, N = m*g*cos (omega)

Explanation:

We can solve this problem with the help of a free body diagram where we show the respective forces in each one of the axes, y & x. The free-body diagram and the equations are in the image attached.

If the product of mass by acceleration is zero, we must clear the normal force of the equation obtained. The acceleration is equal to zero because there is no movement on the Y-axis.

3 0
3 years ago
An object (initially at rest) is dropped from a height h above the ground if it takes 4 seconds for the object to reach the grou
seraphim [82]
Since U=0,
h=1/2gt^2 (h= ut+1/2gt^2, U=0)
h=1/2*10*4*4
h=80m
8 0
3 years ago
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