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ivanzaharov [21]
4 years ago
12

Question 2.

Physics
1 answer:
choli [55]4 years ago
3 0

Explanation:

1 inch = 25.4 mm

1 foot = 12 inches

1 mile = 5260 feet

1 cm = 0.01 m or 10 mm

Now Sammy's height is 5 feet and 5.3 inches.

(a) We need to find Sammy's height in inches.

Since, 1 foot = 12 inches

5 feet = 5 × 12 inches = 60 inches

Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches

Sammy's height is 65.3 inches.

(b) We need to find Sammy's height in feet.

Since, 1 foot = 12 inches

1\ \text{inch}=\dfrac{1}{12}\ \text{feet}

So,

5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}

5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet

Sammy's height is 5.44 feet.

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On top of a cliff of height h, a spring is compressed 5m and launches a projectile perfectly horizontally with a speed of 75 m s
omeli [17]

Answer:

The height of the cliff is 121.276 m

Explanation:

Given;

initial velocity of the projectile, v₁ = 75 m/s

final velocity of the projectile, v₂ = 90 m/s

spring compression = 5 m

Apply the law of conservation of energy;

mgh₀ + ¹/₂mv₁² = mgh₂ + ¹/₂mv₂²

gh₀ + ¹/₂v₁² = gh₂ + ¹/₂v²

gh₁  - gh₂ = ¹/₂v₂² - ¹/₂v₁²

g(h₀  - h₂) = ¹/₂ (v₂² - v₁²)

h₀  - h₂ = ¹/₂g (v₂² - v₁²)

h₀ = h(cliff) + 5m

when the projectile hits the ground, Final height, h₂ = 0

h_o - 0 = \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff} + 5= \frac{1}{2g}(v_2^2-v_1^2)\\\\h_{cliff}  = \frac{1}{2g}(v_2^2-v_1^2) - 5\\\\h_{cliff}  = \frac{1}{2*9.8}(90^2-75^2) - 5\\\\h_{cliff} = 121.276 \ m

Therefore, the height of the cliff is 121.276 m

7 0
3 years ago
What is the mass of an object traveling at 20m/s and having a momentum of 60kg⋅m/s?
vichka [17]

Answer:

m=3kg

Explanation:

Momentum=mass×speed

60[kg m/s]=m[kg]×20[m/s]

We get:

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4 years ago
Read 2 more answers
Can someone please help me think of an idea for this. My teacher wants me to think of a gadget that uses two simple machines, an
ohaa [14]
A dishwasher and a dryer im not sure if that is right but that’s what I would put
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3 years ago
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A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from res
Bumek [7]

Answer:

Part a)

H = 26.8 m

Part b)

error = 7.18 %

Explanation:

Part a)

As the stone is dropped from height H then time taken by it to hit the floor is given as

t_1 = \sqrt{\frac{2H}{g}}

now the sound will come back to the observer in the time

t_2 = \frac{H}{v}

so we will have

t_1 + t_2 = 2.42

\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42

so we have

\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42

solve above equation for H

H = 26.8 m

Part b)

If sound reflection part is ignored then in that case

H = \frac{1}{2}gt^2

H = \frac{1}{2}(9.81)(2.42^2)

H = 28.7 m

so here percentage error in height calculation is given as

percentage = \frac{28.7 - 26.8}{26.8} \times 100

percentage = 7.18

5 0
3 years ago
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