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Tanya [424]
3 years ago
5

You are pulling a sled using a horizontal rèpe, as shown in the diagram. The rope pulls the sled. exerting a force of 50 N to th

e right. The snow exerts a friction force of 30 N on the sled to the left. The mass of the sled is 50 kg.
Find the sun of the force on the self

Determine the acceleration of the sled

If the sled has an initial velocity 2m/s to the right, how fast will it be traveling after 5 seconds?
Physics
1 answer:
geniusboy [140]3 years ago
7 0

Answer:

Part 1

20 N

Part 2

0.4 m/s²

Part 3

4 m/s

Explanation:

The force which pulls the sled right = 50 N

The friction force exterted towards left by the snow = -30 N

The mass of the sled = 50 kg

Part 1

The sum of the forces on the sled, F = 50 N + (-30) N = 20 N

Part 2

The acceleration of the sled is given as follows;

F = m·a

Where;

m = The mass of the sled

a = The accelertion

a = F/m

∴ a = (20 N)/(50 kg) = 0.4 m/s²

The acceleration of the sled, a = 0.4 m/s²

Part 3

The initial velocity of the sled, u = 2 m/s

The kinematic equation of motion to determine the speed of the sled is v = u + a·t

The speed, <em>v</em>, of the sled after t = 5 seconds is therefore;

v = 2 m/s + 0.4 m/s² × 5 s = 4 m/s.

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Two uniformly charged conducting plates are parallel to each other. They each have area A. Plate #1 has a positive charge Q whil
Marrrta [24]

Answer:5

Explanation:

Given

First Plate has a charge of +Q  and area A

Second Plate has a charge of -3 Q and area A

We Know electric Field due to sheet charge is given by

E=\frac{Q}{2A\epsilon }

Where Q=charge over the Plate

A=Area of plate

\epsilon= Permittivity of free space

Electric Field Due to Positive charge will always be away from it while for negative charge it is towards it.

Net Electric Field at a point between between the Plates is the superimposition of two electric with direction

E_{net}=\frac{Q}{2A\epsilon }+\frac{3Q}{2A\epsilon }

E_{net}=\frac{2Q}{A\epsilon }

Net electric Field is towards the negative charged plate

5 0
4 years ago
The design of a 60.0 cm industrial turntable requires that it has a kinetic energy of 0.250 j when turning at 45.0 rpm. What mus
Aneli [31]

Answer:

The moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

Explanation:

Given;

radius of the turnable, r = 60 cm = 0.6 m

rotational kinetic energy, E = 0.25 J

angular speed of the turnable, ω = 45 rpm

The rotational kinetic energy is given as;

E_{rot} = \frac{1}{2} I \omega ^2

where;

I is the moment of inertia about the axis of rotation

ω is the angular speed in rad/s

\omega = 45 \frac{rev}{\min} \times \frac{2 \pi \ rad}{1 \ rev} \times \frac{1 \ \min}{60 \ s} \\\\\omega = 4.712 \ rad/s

E = \frac{1}{2} I \omega ^2\\\\I = \frac{2E}{\omega ^2} \\\\I = \frac{2 \ \times \ 0.25}{(4.712)^2} \\\\I = 0.0225 \ kg.m^2

Therefore, the moment of inertia of the turntable about the rotation axis is 0.0225 kg.m²

5 0
3 years ago
A horizontal force of 100 N is used to pull a crate, which weighs 500 N, at constant velocity across a horizontal floor. What is
Ilya [14]

Answer:

0.2

Explanation:

Horizontal force=100N

Weight of crate=500 N

We have to find the coefficient of kinetic friction.

Normal ,N=Weight=500N

Horizontal force,F_x=\mu_kN

Where F_x=Horizontal force

N=Normal force

\mu_k=Coefficient of kinetic friction

Substitute the values in the formula

100=\mu_k(500)

\mu_k=\frac{100}{500}=0.2

Hence, the coefficient of kinetic friction =0.2

8 0
3 years ago
Two charged particles, q1 and q2, are located on the x-axis, with q1 at the origin and q2 initially atx1 = 14.7 mm.In this confi
Sladkaya [172]

Answer:

The force magnitude is 1.75 μN acting outward from the origin towards x₂

Explanation:

The given parameters, are;

The location of q₁ = The origin

The location of q₂ = 14.7 mm from q₁

The repulsive force exerted on q₂ by q₁ = 2.62 μN

The location the particle q₂ is located to =  18.0 mm from q₁

By Coulomb's law, we have;

F = k\dfrac{q_1 \times q_2 }{r^2}

Where;

k = Coulomb constant ≈ 8.99 × 10⁹ kg·m³/(s²·C²)

r = The distance between the particles

F = The force acting between the particles

When r = 14.7 mm F = 2.62 μN

∴ q₂ × q₁ = r² × F/k = (14.7×10^(-3))²×2.62×10^(-6)/(8.9875517923^9) ≈ 1.48 × 10⁻¹⁸

q₂ × q₁ = 1.48 × 10⁻¹⁸ C²

When the distance is increased to 18.0 mm, we have;

F = (8.9875517923^9) × 1.4796647 × 10^(-18)/((18×10^(-3))²) = 1.75 × 10⁻⁶ N

∴ F = 1.75 μN.

Therefore;

The force magnitude is 1.75 μN outward from the origin towards  x₂.

8 0
4 years ago
If a polythene rod is rubbed with a dry cloth, what kind of charge would it gain? A) positive, due to extra protons. B) positive
Likurg_2 [28]

Answer:

D) negative, due to extra electrons.

Explanation:

The charge that would be gained will be negative charges due to extra electrons.

Electrons are usually lost or gained by bodies that comes in contact with one another.

They occupy the orbital space in an atom and are not strongly held by forces within the atom.

Protons cannot be lost in such manner. They occupy the nucleus and are bounded by strong chemical forces within it.

The polythene and rods are both made up of chemical substances whose units called atoms are made up of subatomic particles of protons, neutrons and electrons. As with all kinds of matter, there would be pool of free electrons round them that can easily be rubbed off due to the weak attractive forces binding them in the atomic sphere.

When the polythene and rods are rubbed, there would be a loss of electrons and the gaining body, polythene becomes negatively charged.

6 0
3 years ago
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