A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block

Answers

Answer 1

Answer:

Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77

What is Friction?

According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.

F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.

Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.

To learn more about Friction, refer to the link:

brainly.com/question/13000653

#SPJ2

Answer 2

Answer:

Answer:

7.77

Explanation:

F=μ*m

n=w which also means n=mg

14.7=0.193*n

n=76.2

76.2=m*9.8

m=7.77

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If the velocity of a pitched ball has a magnitude of 41.0 m/s and the batted ball's velocity is 50.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answers

The change in momentum is (91 m/s) multiplied by the mass of the ball (which you neglected to mention).
That's exactly the impulse delivered by the bat.

A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potential at this surface?

Answers

Answer:

V = 575.6 Volts

Explanation:

As we know that surface area of the equi-potential surface is given as

$A = 1.18 m^2$

so we will say

$A = 4\pi r^2$

$1.18 = 4\pi r^2$

$r = 0.31 m$

Now the potential due to a point charge is given as

$V = (kQ)/(r)$

$V = ((9* 10^9)(1.96 * 10^(-8)))/(0.31)$

$V = 575.6 Volts$

Hercules X-1 is a pulsating X-ray source. The X-rays from this source sometimes completely disappear for 6 hours every 1.7 days because the neutron star has a 1.7-day orbital period around its companion star, and it is eclipsed for ____ hours once every orbital period.

Answers

Answer:

06 Hours

Explanation:

As per the details given in the question it self, the neutron star X-1 is revolving around its companion star. The orbital period is 1.7 years which means it will complete the revolution in 1.7 years. During the movement in the orbit we will be able to detect the x-rays except for the time when it goes behind the companion star and eclipsed by it as seen from Earth.

Since the x-rays disappear completely for around 6 hours. This clearly means that eclipse period is 06 hours.

A flat (unbanked) curve on a highway that has a radius of 50 m. A car rounds the curve. The car has mass 4,907 kg. The static coefficient of friction between the curve and the car is 0.35. What is the maximum speed of the car to prevent sliding?

Answers

Answer:

maximum speed of the car to prevent sliding is 13.1m/s

Explanation:

Given data

Radius of curve r=50m

Mass of car m=4907kg

Coefficient of friction u=0.35

Limiting for R=?

Hence limiting force R=ma

R=4907*9.81

R=48137.7N

We know that the force to overcome friction is

F=uR

Hence

F=0.35*48137.7

F=16848.2N

Centripetal force along the curve is given as

Fc=mv²/r

Fc = centripetal force

m = mass

v = velocity

r = radius

To solve for velocity we have to equate both force required to overcome friction and the centripetal force

Fc=mv²/r=F=uR

mv²/r=uR

Making velocity subject of formula we have

v²=u*r*R/m

v²=(0.35*50*48137.7)/4907

v²=842409.75/

v²=171.67

v=√171.67

v=13.1m/s

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Answers

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(b) Calculate the magnification.

Answers

Answer:

A) Calculate the distance

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