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White raven [17]
3 years ago
5

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the

object sliding at a constant velocity. The magnitudes of the required forces are different in these situations because the force of kinetic friction
a) decreases as the speed of the object relative to the surface increases.

b)is greater than the force of static friction.

c) increases as the speed of the object relative to the surface increases.

d) is less than the force of static friction
Physics
1 answer:
taurus [48]3 years ago
3 0

The force required to start an object sliding across a uniform horizontal surface is larger than the force required to keep the object sliding at a constant velocity once it starts.

The magnitudes of the required forces are different in these situations because the force of kinetic friction is less than the force of static friction. <em>(d)</em>

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What do protoplanets eventually grow into?​
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Answer:

a planet

Explanation:

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7 0
2 years ago
What is the half-life of a radioactive isotope that decreases to one fourth its original amount in 18 months
just olya [345]
9 months, because half of 18 is 9
7 0
3 years ago
A wire 6.90 m long with diameter of 2.15 mm has a resistance of 0.0320 Ω. Find the resistivity of the material of the wire. rho
spayn [35]
<h2>Answer:</h2>

1.68 x 10⁻⁸Ωm

<h2>Explanation:</h2>

The resistance (R) of a wire is related to its length(L), its material resistivity(ρ) and its crossectional area(A) as follows;

R = ρL/A               ------------------------(i)

Where;

A = πd² / 4              [where d = diameter of the wire]

From the question;

L = 6.90m

d = 2.15mm = 0.00215m

R = 0.0320Ω

First calculate the crossectional area (A) of the wire as follows;

A = πd² / 4      

[Take π = 3.142]

d = 0.00215m

∴ A = 3.142 x (0.00215)² / 4

∴ A = 0.000003631m²

Now, substitute the values of A, L, and R into equation (i) as follows;

R = ρL/A

0.0320 = ρ x 6.90 / 0.000003631

0.0320 = 1900302.95 x ρ

Solve for ρ;

=> ρ = 0.0320 / 1900302.95

=> ρ = 1.68 x 10⁻⁸Ωm

Therefore, the resistivity of the material of the wire is 1.68 x 10⁻⁸Ωm

4 0
3 years ago
Light shines through a single slit whose width is 5.7 x 10-4 m. A diffraction pattern is formed on a flat screen located 4.0 m a
lilavasa [31]

Answer:

\lambda = 570\ nm

Explanation:

Given,

Width of slit, W = 5.7 x 10⁻⁴ m

Distance between central bright fringe, L = 4 m

distance between central bright fringe and first dark fringe, y = 4 mm

Diffraction angle

tan \theta = \dfrac{y}{L}

tan \theta = \dfrac{4}{4\times 10^3}

\theta = 0.0572

Now.

W sin \theta = m \lambda

m = 1

5.7 \times 10^{-4} \times sin (0.0572) = 1 \times \lambda

\lambda = 569.99 \times 10^{-9}\ m

\lambda = 570\ nm

4 0
4 years ago
How would you differentiate the theoretical and actual velocities of PE and KE in falling objects?
Solnce55 [7]

Answer: you've been trolled

Explanation:

8 0
3 years ago
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