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solong [7]
3 years ago
14

PLEASE HELP ME GUYS IM DESPERATE!!!

Physics
1 answer:
geniusboy [140]3 years ago
8 0
Number 1 is D number 2 is C
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A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.
dsp73

Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

\mu = \frac{F_f}{R}

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta

Given

θ = 24°

\mu = tan 24^0\\\mu = 0.445\\

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0
3 years ago
What is the direction of the electric force exerted by paperclip 1 on paperclip 2?
Tema [17]
In general, the electric force in an electric field is exerted outward from a positive atom, and inward for the negative atom. Therefore based on the figure you specified, we can say that the <span>electric force exerted by paperclip 1 on paperclip 2 is repulsive.</span>
6 0
4 years ago
How do you rationalize the tension being used in Tennis Racket strings using the concept of impulse and momentum?
zheka24 [161]

Answer:

The momentum, ΔP, and therefore, kinetic energy given to the ball in a serve is the result of the product of the tension force, 'F', in the string and the time of contact, Δt, between the ball and the string

ΔP = F × Δt

Explanation:

The impulse, ΔP, is the produce of the force, 'F', applied to a body for a given period of time, Δt', that gives motion to the body, and it is equal to the change of momentum of the body

ΔP = F × Δt

The momentum, 'P', of a body is the product of the mass, 'm', of the body and its velocity, 'v'

P = m × v

Tension is the axial pulling force of a string

T = Axial Force, F_{axial}

The tension used in Tennis Racket strings is between 40 to 65 lbs.

When high tension is used in the string, the string is taut, and the contact duration between the Racket string and the ball is minimal, and the player needs to use more force to obtain a high momentum, and therefore, energy in the ball, which reduces control, and increase stress, as force is more emphasized

When low tension is used in the string, the Tennis Racket strings are more elastic. During a serve, the ball pushes the strings further back into the racket, such that the ball spends more time in contact with the string, (Δt is larger), and therefore, the impulse, F·Δt = ΔP, given to the ball is larger, therefore, the ball has a larger change in momentum, and therefore more energy in the intended direction.

However, a very slackened string will increase the increase area and time (large Δt) of contact of the ball and the racket such that the force given to the ball, F = ΔP/(large Δt) is reduced and therefore reduce the likelihood of gaining points from a serve against an opponent with a much forceful return of a serve.

3 0
3 years ago
A hollow cylinder of mass 2.00 kgkg, inner radius 0.100 mm, and outer radius 0.200 mm is free to rotate without friction around
kipiarov [429]

Answer: 2.86 m

Explanation:

To solve this question, we will use the law of conservation of kinetic and potential energy, which is given by the equation,

ΔPE(i) + ΔKE(i) = ΔPE(f) + ΔKE(f)

In this question, it is safe to say there is no kinetic energy in the initial state, and neither is there potential energy in the end, so we have

mgh + 0 = 0 + KE(f)

To calculate the final kinetic energy, we must consider the energy contributed by the Inertia, so that we then have

mgh = 1/2mv² + 1/2Iw²

To get the inertia of the bodies, we use the formula

I = [m(R1² + R2²) / 2]

I = [2(0.2² + 0.1²) / 2]

I = 0.04 + 0.01

I = 0.05 kgm²

Also, the angular velocity is given by

w = v / R2

w = 4 / (1/5)

w = 20 rad/s

If we then substitute these values in the equation we have,

0.5 * 9.8 * h = (1/2 * 0.5 * 4²) + (1/2 * 0.05 * 20²)

4.9h = 4 + 10

4.9h = 14

h = 14 / 4.9

h = 2.86 m

8 0
3 years ago
Read 2 more answers
Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.1×104 m/s when at a distance
trapecia [35]

Answer:

v_f = 6.92 x 10^(4) m/s

Explanation:

From conservation of energy,

E = (1/2)mv² - GmM/r

Where M is mass of sun

Thus,

E_i = E_f will give;

(1/2)mv_i² - GmM/(r_i) = (1/2)mv_f² - GmM/(r_f)

m will cancel out to give ;

(1/2)v_i² - GM/(r_i) = (1/2)v_f² - GM/(r_f)

Let's make v_f the subject;

v_f = √[(v_i)² + 2MG((1/r_f) - (1/r_i))]

G is Gravitational constant and has a value of 6.67 x 10^(-11) N.m²/kg²

Mass of sun is 1.9891 x 10^(30) kg

v_i = 2.1×10⁴ m/s

r_i = 2.5 × 10^(11) m

r_f = 4.9 × 10^(10) m

Plugging in all these values, we have;

v_f = √[(2.1×10⁴)² + 2(1.9891 x 10^(31)) (6.67 x 10^(-11))((1/(4.9 × 10^(10))) - (1/(2.5 × 10^(11)))] 20.408 e12

v_f = √[(441000000) + 2(1.9891 x 10^(30)) (6.67 x 10^(-11))((16.408 x 10^(-12))]

v_f = √[(441000000) + (435.38 x 10^(7))

v_f = 6.92 x 10^(4) m/s

5 0
4 years ago
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