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Andru [333]
3 years ago
12

A person applies a force of 200N to move an object initially at rest weighing 400N. What could be the coefficient of static fric

tion between the object and the surface?
Physics
1 answer:
SOVA2 [1]3 years ago
5 0

Answer:

\mu=0.5

Explanation:

Given that,

Applied force, F = 200 N

Weight of the object, or the normal force, N = 400 N

We need to find the coefficient of static friction between the object and the surface.

The force of friction has the same magnitude as that of applied force. It can be given by :

F=\mu mg

mg is weight of the object

\mu=\dfrac{F}{mg}\\\\\mu=\dfrac{200}{400}\\\\\mu=0.5

So, the coefficient of static friction between the object and the surface is 0.5.

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A car moves at speed v across a bridge made in the shape of a circular arc of radius r. (a) Find an expression for the normal fo
inna [77]

Answer:

(a) FN = m (g - \frac{v^{2} }{r})

(b) vmin = 17.146 m/s

Explanation:

The radius of the arc is

r = 30m

The normal force acting on the car form the highest point is

FN = m (g - \frac{v^{2} }{r})

If the normal force become 0 we have

m (g - \frac{v^{2} }{r}) = 0

or

g - \frac{v^{2} }{r} = 0

This way, when FN = 0, then v = vmin, so

g - \frac{vmin^{2} }{r} = 0

vmin = \sqrt[.]{g*r} = \sqrt[.]{9.8 m/s^{2} * 30m } = 17.146 m/s

4 0
3 years ago
Your friend has decided to make some money during the next State Fair by inventing a game of skill. In the game as she has devel
coldgirl [10]

Answer: from the vertical, one should aim  86.6°

Explanation:

height of the center of object = 7.0 m - 0.05 m = 6.95 m

now let the bullet hits centre at point A height x meters from the ground

also let t be the time taken for the bullet to hit the object

so distance travelled by the target will be

d = h - x = 6.95 - x

now using the equation of motion

d = 1/2gt²

so 1/2gt² = 6.95 - x

x = 6.95 - 1/2gt² .........let this be equ 1

let angle of fire be ∅

so v(cos∅) × t = 100

our velocity v is 1200 ft/sec = 365.76 m/s

365.76(cos∅) × t = 100 ........equ 2

also vertical position of the bullet after t is

y = y₀ + c(sin∅)t - 1/2gt²

y = 1 + 365.76(sin∅)t - 1/2gt² ----- equ 3

After time t. the vertical position x and y are same, else the bullet wouldn't have strike target at centre, so;

x = y

we substitute

equ 1 = equ 3

6.95 - 1/2gt² = 1 + 365.76(sin∅)t - 1/2gt²

6.95 - 1 = 365.76(sin∅)t - 1/2gt² +  1/2gt²

5.95 = 365.76(sin∅)t

t = 5.96 / 365.76(sin∅)

now input the above equ  into equ 2

365.76(cos∅) ×  5.96/365.76(sin∅) = 100

5.95(cos∅)/sin∅ = 100

tan∅ = 5.95/100 = 0.0595

∅ =  3.40°

therefore from the vertical, one should aim (90° - 3.40°) = 86.6°

4 0
3 years ago
a motorcycle accelerates from 15 m/s to 20 m/s over a distance of 50 meters. what is its average acceleration?
devlian [24]
For this, you need the v-squared equation, which is v(final)² = v(initial)² + 2aΔx The averate acceleration is thus a = (v(final)² - v(initial)²) / 2Δx = (20² - 15²) / 2(50) = 175 / 100 = 1.75 m/s² So the average acceleration is 1.75 m/s²
5 0
3 years ago
A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull
sdas [7]

In spring mass system we know that angular frequency is given as

\omega = 2\pi f

f = 8.38 Hz

\omega = 2\pi(8.38)

\omega = 52.65  rad/s

now we know that speed of SHM at its extreme position is given by

v = A\omega

here we know that

A = 17.5 cm

v = 0.175 (52.65)

v = 9.21 m/s

so maximum speed is 9.21 m/s

7 0
3 years ago
How much heat energy is required to raise the temperature of 0.368kg of copper from 23.0 ∘C to 60.0 ∘C? The specific heat of cop
Katarina [22]
The change in temperature here corresponds to a sensible heat. The amount of energy required can be calculated by multiplying the specific heat capacity, the amount of the substance and the corresponding change in temperature.

Heat required = mCΔT
Heat required = 0.368 kg (0.0920 cal/g°C) (60 - 23)°C
Heat required = 1.25 cal
3 0
3 years ago
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