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2 years ago

5

During summer break, Jamie's family went camping in the mountains. There were no roads to their camp, 8 miles away, and it took

them about 4.5 hours to reach the campsite on foot. Did Jamie's family stop to rest during the trip? How can you tell from the data?
A) No. They continued moving but their speed varied.
B) Yes. After 1.5 hours, there is no change in position for 30 minutes.
C) No. They continued moving, although they also slowed down continuously during the trip.
D) Not really. They did stop, but that was only when they reached the campsite at distance equals 0.

2 answers:

liraira [26]2 years ago

5 0

The Answer is B

hope this helps

Rama09 [41]2 years ago

4 0

Answer:

B)Yes. After 1.5 hours, there is no change in position for 30 minutes.

Explanation:

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Two billiard balls of equal mass move at right angles and meet at the origin of an xy coordinate system. Initially ball A is mov

frez [133]

Answer:

Speed of ball A after collision is 3.7 m/s

Speed of ball B after collision is 2 m/s

Direction of ball A after collision is towards positive x axis

Total momentum after collision is m×4·21 kgm/s

Total kinetic energy after collision is m×8·85 J

Explanation:

<h3>If we consider two balls as a system as there is no external force initial momentum of the system must be equal to the final momentum of the system</h3>

Let the mass of each ball be m kg

v $_{1}$ be the velocity of ball A along positive x axis

v $_{2}$ be the velocity of ball A along positive y axis

u be the velocity of ball B along positive y axis

Conservation of momentum along x axis

m×3·7 = m× v $_{1}$

∴ v $_{1}$ = 3.7 m/s along positive x axis

Conservation of momentum along y axis

m×2 = m×u + m× v $_{2}$

2 = u + v $_{2}$ → equation 1

<h3>Assuming that there is no permanent deformation between the balls we can say that it is an elastic collision</h3><h3>And for an elastic collision, coefficient of restitution = 1</h3>

∴ relative velocity of approach = relative velocity of separation

-2 = v $_{2}$ - u → equation 2

By adding both equations 1 and 2 we get

v $_{2}$ = 0

∴ u = 2 m/s along positive y axis

Kinetic energy before collision and after collision remains constant because it is an elastic collision

Kinetic energy = (m×2² + m×3·7²)÷2

= 8·85×m J

Total momentum = m×√(2² + 3·7²)

= m× 4·21 kgm/s

3 0

2 years ago

A weather balloon is rising vertically from a launching pad on the ground. A technician standing 300 feet from the launching pad

avanturin [10]

Answer:

$\dfrac{dh}{dt} =5\ ft/s$

Explanation:

Let

h = height of balloon (in feet).

θ = angle made with line of sight and ground (in radians).

h = 300 tanθ

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

now $\dfrac{dh}{dt}$ can be written as

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{d\theta }{dt} = \dfrac{1}{120}\at \ \theta =\dfrac{\pi}{4}$

When θ = π/4,

$\dfrac{dh}{d\theta } = 300 sec^2\theta$

$\dfrac{dh}{d\theta } = 600$

$\dfrac{dh}{dt} =\dfrac{dh}{d\theta }\times \dfrac{d\theta }{dt}$

$\dfrac{dh}{dt} =600\times \dfrac{1}{120}$

$\dfrac{dh}{dt} =5\ ft/s$

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2 years ago

a 75 kg man is standing at rest on ice while holding a 4kg ball. if the man throws the ball at a velocity of 3.50 m/s forward, w

AysviL [449]

Answer:

His resulting velocity will be 0.187 m/s backwards.

Explanation:

Given:

Mass of the man is, $M=75\ kg$

Mass of the ball is, $m=4\ kg$

Initial velocity of the man is, $u_m=0\ m/s(rest)$

Initial velocity of the ball is, $u_b=0\ m/s(rest)$

Final velocity of the ball is, $v_b=3.50\ m/s$

Final velocity of the man is, $v_m=?\ m/s$

In order to solve this problem, we apply law of conservation of momentum.

It states that sum of initial momentum is equal to the sum of final momentum.

Momentum is the product of mass and velocity.

Initial momentum = Initial momentum of man and ball

Initial momentum = $Mu_m+mu_b=75\times 0+4\times 0 =0\ Nm$

Final momentum = Final momentum of man and ball

Final momentum = $Mv_m+mv_b=75\times v_m+4\times 3.50 =75v_m+14$

Now, initial momentum = final momentum

$0=75v_m+14\\\\75v_m=-14\\\\v_m=\frac{-14}{75}\\\\v_m=-0.187\ m/s$

The negative sign implies backward motion of the man.

Therefore, his resulting velocity is 0.187 m/s backwards.

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2 years ago

Two parallel co-axial disks are floating in deep space (far from sun and planets). Each disk is 1 meter in diameter and the disk

HACTEHA [7]

Answer:

T₂ = 5646 K

Explanation:

Let's start by finding the power received by the first disc, for this we use Stefan's law

P = σ. A e T⁴

Where next is the Stefam-Bolztmann constant with value 5,670 10-8 W / m² K⁴, A is the area of the disk, T the absolute temperature and e the emissivity that for a black body is 1

The intensity is defined as the amount of radiation that arrives per unit area. For this we assume that the radiation expands uniformly in all directions, the intensity is

I = P / A

Writing this expression for both discs

I₁ A₁ = I₂ A₂

I₂ = I₁ A₁ / A₂

The area of a sphere is

A = 4π r²

I₂ = I₁ (r₁ / r₂)²

r₂ = r₁ ± 5

I₁ = I₂ ( (r₁ ± 5)/r₁)²

.

Let's write the Stefan equation

P / A = σ e T⁴

I = σ e T⁴

This is the intensity that affects the disk, substitute in the intensity equation

σ e₁ T₁⁴ = σ e₂ T₂⁴ (r₂ / r₁)²

The first disc indicates that it is a black body whereby e₁ = 1, the second disc, as it is painted white, the emissivity is less than 1, the emissivity values of the white paint change between 0.90 and 0.95, for this calculation let's use 0.90 matt white

e₁ T₁⁴ = T₂⁴ (r1 + 5)²/r₁²

T₁ = T₂ {(e₂/e₁)}^{1/4} √(1 ± 1/ r₁)

If we assume that r₁ is large, which is possible since the disks are in deep space, we can expand the last term

(1 ±x) n = 1 ± n x

Where x = 5 / r₁ << 1

We replace

T₁ = T₂ {(e₂/e₁)}^{1/4} (1 ± ½ 5/r1)

T₁ = T₂ {(e₂)}^{1/4} (1 ± 5/2 1/r1)

If the discs are far from the star, they indicate that they are in deep space, the distance r₁ from being grade by which we can approximate; this is a very strong approach

T₁ = T₂ {(e₂)}^{1/4} ¼

T<u>₁</u> = T₂ 0.9 $0.9^{1/4}$

5500 = T₂ 0.974

T₂ = 5646 K

3 0

2 years ago

Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

schepotkina [342]

If speed = distance/time , then time = speed/distance.

So...

Speed of light = 3*10^8(m/s)
Average distance from Earth to Sun = 149.6*10^9(m)

Therefore, t=(3*10^8(m/s))/(149.6*10^9(m))

I hope this was a helpful explanation, please reply if you have further questions about the problem.

Good luck!

5 0

2 years ago