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dimulka [17.4K]
3 years ago
14

A 62 pound box is on an incline. Determine the minimum force P that would result in the box starting to slide up the incline. (i

nclude units with answer)
Physics
1 answer:
Anon25 [30]3 years ago
8 0

Answer:

F > W * sin(α)

Explanation:

The force needed for the box to start sliding up depends on the incline (α).

The external forces acting on the box would be the weight, the normal reaction and the lifting force that is applied to make it slide up.

These forces can be decomposed on their normal and tangential (to the slide plane) components.

The weight will be split into

Wn = W * cos(α) (in normal direction)

Wt = W * sin(α) (in tangential direction)

The normal reaction will be alligned with the normal axis, and will be equal to -Wn

N = -W* cos(α) (in normal direction)

To mke the box slide up, a force must be applied, that is opposite to the tangential component of the weight and at least a little larger

F > |-W * sin(α)| (in tangential direction)

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A(n) 82.7 kg boxer has his first match in the Canal Zone with gravitational acceleration 9.782 m/s 2 and his second match at the
Annette [7]

Answer:

82.7 kg

Explanation:

the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).

6 0
3 years ago
A 77.0−kg short-track ice skater is racing at a speed of 12.6 m/s when he falls down and slides across the ice into a padded wal
sukhopar [10]

Answer:

-6112.26  J

Explanation:

The initial kinetic energy, KE_i is given by

KE_i=0.5mv_1^{2} where m is the mass of a body and v_i is the initial velocity

The final kinetic energy, KE_f is given by

KE_f=0.5mv_f^{2} where v_f is the final velocity

Change in kinetic energy, \triangle KE is given by

\triangle KE=KE_f-KE_i=0.5mv_f^{2}-0.5mv_1^{2}=0.5m(v_f^{2}-v_i^{2})

Since the skater finally comes to rest, the final velocity is zero. Substituting 0 for v_f and 12.6 m/s for v_i and 77 Kg for m we obtain

\triangle KE=0.5*77*0^{2}-0.5*77*(0^{2}-12.6^{2})=-6112.26 J

From work energy theorem, work done by a force is equal to the change in kinetic energy hence for this case work done equals <u>-6112.26  J</u>

3 0
3 years ago
Q1: An object with a charge of 1.2 C is located 4.5 m away from a second object that has a charge of 0.36 C. Find the electrical
gizmo_the_mogwai [7]

Answer:

a) F= 0,19  [N]   according to problem statement

b) F = 0,19*10⁹ [N]  using the right value of K

Explanation:

The force between two electric charges is according to Coulomb´s law is:

F = K * q₁*q₂ / d²    where  q₁  and q₂ are the charges on body one and body 2 respectively, d is the distance between the two bodies and K is a constant  K = 8,988100*10⁹ N.m²/C². The problem establishes to use        K = 8,988100 N.m²/C².

NOTE: To value of is :  K = 8,988100*10⁹ N.m²/C². I am going to solve the problem using K = 8,988100 N.m²/C² if that information was an error, all we need to get the right answer is multiply the result by 10⁹

Then:

F = 8,988100 * 1,2* 0,36 / (4,5)²     [ N*m²/C² ] * [ C*C*/m²]

F = 3,882859/ 20,25  [N]

F= 0,19  [N]

The force is of repulsion since the two charges are positive and in the direction of the straight line which passes through the centers of the bodies

4 0
3 years ago
A 50-kg toboggan is coasting on level snow. As it passes beneath a bridge, a 20-kg parcel is dropped straight down and lands in
timama [110]

Answer:

KE2/KE1=0.71

Explanation:

By conservation of the linear momentum:

m1*V1 = (m1+m2)*V2

Solving for V2:

V2 = \frac{m1}{m1+m2}*V1

The kinetic energies are:

KE1=1/2*m1*V1^2

KE2 = 1/2*(m1+m2)*V2^2

KE2 = 1/2*(m1+m2)*(\frac{m1}{m1+m2}*V1)^2

Simplifying:

KE2 = 1/2*\frac{m1^2}{m1+m2}*V1^2

The ratio will be:

KE2/KE1=\frac{1/2*\frac{m1^2}{m1+m2}*V1^2}{1/2*m1*V1^2} =\frac{m1}{m1+m2}

KE2/KE1=0.71

5 0
3 years ago
A stretched string has a mass per unit length of 4.87 g/cm and a tension of 16.7 N. A sinusoidal wave on this string has an ampl
Burka [1]

Answer:

Explanation:

Given that,

Mass per unit length is

μ = 4.87g/cm

μ=4.87g/cm × 1kg/1000g × 100cm/m

μ = 0.487kg/m

Tension

τ = 16.7N

Amplitude

A = 0.101mm

Frequency

f = 71 Hz

The wave is traveling in the negative direction

Given the wave form

y(x,t) = ym• Sin(kx + ωt)

A. Find ym?

ym is the amplitude of the waveform and it is given as

ym = A = 0.101mm

ym = 0.101mm

B. Find k?

k is the wavenumber and it can be determined using

k = 2π / λ

Then, we need to calculate the wavelength λ using

V = fλ

Then, λ = V/f

We have the frequency but we don't have the velocity, then we need to calculate the velocity using

v = √(τ/μ)

v = √(16.7/0.487)

v = 34.29

v = 5.86 m/s

Then, we can know the wavelength

λ = V/f = 5.86 / 71

λ = 0.0825 m

So, we can know the wavenumber

k = 2π/λ

k = 2π / 0.0825

k = 76.18 rad/m

C. Find ω?

This is the angular frequency and it can be determined using

ω = 2πf

ω = 2π × 71

ω = +446.11 rad/s

D. The angular frequency is positive (+) because the direction of propagation of wave is in the negative direction of x

5 0
3 years ago
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