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3 years ago

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A model rocket accelerates upward from the ground with a constant acceleration, reaching a height of 50 m in 8 s.What is the spe

ed (in m/s) at a height of 50 m?A: 1.07×101 B: 1.25×101 C: 1.46×101 D: 1.71×101 E: 2.00×101

1 answer:

LekaFEV [45]3 years ago

5 0

Answer:

The value is $v = 12.5 \ m/s$

Explanation:

From the question we are told that

The height is $h = 50 \ m$

The time taken is $t = 8 \ s$

From the equation of motion we have that

$s = ut + \frac{1}{2} * a * t^2$

Here u = 0 because the rocket started at rest

$50 = 0 + \frac{1}{2} * a * 8^2$

=> $a = \frac{100}{64}$

=> $a = 1.5625 \ m/s^2$

Also from the kinematic equation we have that

$v = u + at$

=> $v = 0 + (8 * 1.5625)$

=> $v = 12.5 \ m/s$

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This graph describes the motion of a sky diver that jumped from an airplane. The skydiver is MOST LIKELY ?

tensa zangetsu [6.8K]

Answer:

A) being influenced by equal amounts of gravity and air resistance.

Explanation:

B) slowing down because of an unbalanced force of air resistance.

False - if it was slowing down, then the velocity would go down.

D) on the ground and is not falling anymore.

False - This would be mistaken as the answer but it is not because if the person is not falling anymore the horizontal line should be at the x-axis, meaning that there is no more velocity.

C) accelerating because of an unbalanced force of gravity.

False - The line would otherwise be going up or down.

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3 years ago

A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. W

Lilit [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The total pressure is $P_T = 10.79*10^{5} N/m^2$

The temperature at the bottom is $T_b = 284.2 \ K$

Explanation:

From the question we are told that

The length of the glass tube is $L = 1.50 \ m$

The length of water rise at the bottom of the lake $d = 1.33 \ m$

The depth of the lake is $h = 100 \ m$

The air temperature is $T_a = 27 ^oC = 27 +273 = 300 \ K$

The atmospheric pressure is $P_a = 1.01 *10^{5} N/m$

The density of water is $\rho = 998 \ kg/m^3$

The total pressure at the bottom of the lake is mathematically represented as

$P_T = P_a + \rho g h$

substituting values

$P_T = 1.01*10^{5} + 998 * 9.8 * 100$

$P_T = 10.79*10^{5} N/m^2$

According to ideal gas law

At the surface the glass tube not covered by water at surface

$P_a V_a = nRT_a$

Where is the volume of

$P_a *A * L = nRT_a$

At the bottom of the lake

$P_T V_b = nRT_b$

Where $V_b$ is the volume of the glass tube not covered by water at bottom

and $T_b$ i the temperature at the bottom

So the ratio between the temperature at the surface to the temperature at the bottom is mathematically represented as

$\frac{T_b}{T_a} = \frac{d * P_T}{P_a * h}$

substituting values

$\frac{T_b}{27} = \frac{0.133 * 10.79 *10^5}{1.01 *10^{5} * 1.5}$

=> $T_b = 284.2 \ K$

3 0

3 years ago

By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each obj

NNADVOKAT [17]

Answer:

Distance between both the object will be 404.51 m

Explanation:

We have given charge on two objects $q_1=q_2=1C$

Coulomb force between the two objects is given F = 5.5 N

We have to fond the distance between both the object so that force between them is 5.5 N

According to coulomb law force between two charge particle is $F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}$

So $5.5=\frac{9\times 10^9\times 1\times 1}{r^2}$

$r^2=16.36\times 10^8$

r = 404.51 m

So distance between both object will be 404.51 m

6 0

3 years ago

a cheetah starting from rest accelerates at a rate of 2.5m/s^2 to reach a speed of 45 m/s.how fare does the cheetah travel

Helen [10]

Answer:

405 m

Explanation:

Given:

v₀ = 0 m/s

a = 2.5 m/s²

v = 45 m/s

Find: Δx

v² = v₀² + 2aΔx

(45 m/s)² = (0 m/s)² + 2 (2.5 m/s²) Δx

Δx = 405 m

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I cant belive im doing this- i will mark brainliest if this gets answered- mostly correct.

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I’m going to send a picture

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