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2 years ago

James has a mass of 98 kg and Basma has a mass of 59 kg. James is running at 3.0 m/s, while Basma is running at 4.0 m/s.

Physics

1 answer:

belka [17]2 years ago

5 0

a. The most momentum is attained by James.

b. The resultant force acting on her is 90.77N.

c. Time taken for him to stop is 3.24 s

<h3>What is momentum?</h3>

Momentum of any object is the product of the velocity of the object and its mass.

P =mv

James has a mass of 98 kg and Basma has a mass of 59 kg. James is running at 3.0 m/s, while Basma is running at 4.0 m/s.

Substitute the values, we get

P jane= 98 kg x 3

P jane= 294 kg.m/s

Momentum of Basma, is

P Basma = 59 kg x 4

P Basma = 236 kg.m/s

Therefore, the momentum is large for James.

b. Basma approaches a road, so she stops quickly. It takes her 2.6 seconds to stop. The resultant force acting on her

F. t = P Basma

F =[ 236/2.6

F = 90.77 N

Thus, the resultant force on Basma is 90.77 N.

c. James also stops for the road. If he were to have the same resultant force acting on him, Time taken for him to stop is

F = 90.77 = P james/t

t = 294 / 90.77

t = 3.24 s

Thus, time taken for James to stop for the same resultant force is 3.24 s

Learn more about momentum.

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Answer:

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A friction force of 280 N exists between a cart and the path. If the force of reaction is 766 N, what is the minimum action forc

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here we also know the reaction force due to surface

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so we can say reaction force is given as

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$766 = \sqrt{F_n^2 + 280^2}$

$F_n^2 = 766^2 - 280^2$

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now by force balance we will say

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$F_y = mg - F_n$

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$F_{min} = \frac{\mu mg}{\sqrt{1 + \mu^2}}$

here we know that

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An airplane is flying at 635 km per hour at an altitude of 35,000 m. What is its velocity?

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2 years ago

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

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To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

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$k = R^2 --- (2)$

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$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

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$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

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