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yanalaym
3 months ago
15

A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate a

nd ramp is 0.22. A horizontal force of 5000 N is applied to the crate, pushing it up the ramp. What is the acceleration of the crate?

Physics
2 answers:
inna [3.1K]3 months ago
6 0

Answer:

8.35 m/s^2

Solution Explanation:

Given data:

Mass of the crate = 250 kg

\theta=30^{\circ}

Kinetic friction coefficient between crate and ramp = 0.22

Applied horizontal force = 5000 N

Objective: Calculate the crate's acceleration.

The normal force acting perpendicular to the ramp surface is:

=N=Fsin\theta+mgcos\theta=5000sin 30^{\circ}+250\times 9.8\times cos30^{\circ}=4623.93 N

The friction force opposing the crate's motion is calculated as:

0.22\times 4623.93=1017.26 N

According to Newton's second law, the net force causing the acceleration is:

ma=Fcos\theta-(F_f+mgsin\theta)

a=\frac{5000cos 30^{\circ}-(1017.26+250\times 9.8 sin30^{\circ}}{250}

a=8.35 m/s^2

Therefore, the crate’s acceleration is: 8.35 m/s^2

serg [3.5K]3 months ago
6 0

The force moving the crate upward along the incline acts against at least two other forces: gravitational force pulling it downwards and friction between the crate and the ramp’s rough surface.

Force on any object can be described through its mass and acceleration.

The kinetic friction force is determined by the relation between applied force and friction force, taking gravity into account.

So, 5000 minus 22% of 5000 equals 250 multiplied by acceleration.

Therefore, acceleration equals 3900 divided by 250 meters per second squared.

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Two charges of magnitude 5nC and -2nC are placed at points (2cm,0,0) and
ValentinkaMS [3465]

Answer:

20 cm

Explanation:

The electric potential energy U is calculated with the formula U = kq₁q₂/r, where q₁ = 5 nC (5 × 10⁻⁹ C) and q₂ = -2 nC (-2 × 10⁻⁹ C) and r is determined as √(x - 2)² + (0 - 0)² + (0 - 0)² = x - 2. This leads to U = -0.5 µJ (-0.5 × 10⁻⁶ J), where k = 9 × 10⁹ Nm²/C².

Thus, solving for r gives us r = kq₁q₂/U

which leads to x - 2 = kq₁q₂/U

Then, rearranging gives x = 0.02 + kq₁q₂/U m

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Resulting in x = 0.02 - 90 × 10⁻⁹ Nm²/-0.5 × 10⁻⁶ J

This simplifies to x = 0.02 + 0.18 = 0.2 m, or 20 cm

7 0
2 months ago
Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of
inna [3103]

Answer:

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b) The type of radiation is microwave radiation.

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7 0
3 months ago
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