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sasho [114]
3 years ago
13

Please help with these physics problems 1. A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to

the ground. The height of the building is...? 2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity? 3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?
Physics
1 answer:
Tatiana [17]3 years ago
8 0

Answer:

1.) h = 164.8 m

2.) U = 49.1 m/s

3.) t = 1.43 seconds

Explanation:

1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...? 

Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero

Using second equation of motion

h = Ut + 1/2gt^2

Substitutes the time into the formula

h = 1/2 × 9.8 × 5.8^2

h = 164.8 m

2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?

At maximum height final velocity = 0

Using the third law of motion

V^2 = U^2 - 2gH

0 = U^2 - 2 × 9.8 × 123

U^2 = 2410.8

U = 49.1 m/s

3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?

Since the apple fall from rest, the initial velocity U will be equal to zero

Using the second equation of motion,

h = Ut + 1/2gt^2

substitute all the parameters into the formula

10 = 1/2 × 9.8 × t^2

10 = 4.9t^2

t^2 = 10/4.9

t^2 = 2.04

t = 1.43 seconds

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A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if
diamong [38]

Answer:

I believe that the answer is d.

Explanation:

Because there is nothing to make the aircraft accelerate or decelerate, it is going to stay in constant motion with no acceleration.

3 0
3 years ago
If you see a sedimentary rock outcrop and red layers of sand are on top of pale yellow layers of sand, what do you know for sure
neonofarm [45]

Answer: The yellow layer is definitely older than the red layer

Explanation: According to Nicolaus Steno's law of superposition and original horizontality. Older rocks underlie younger rocks.

Sedimentary rocks are usually deposited in horizontal layers in which each stratigraphic layer is laid down before another can be deposited upon it.

The red layer, in addition to being older, is also likely to have undergone intense oxidation due to earlier exposure.

5 0
3 years ago
Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const
bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

S = ut + \frac{1}{2}at^{2}

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0
2 years ago
Convert 3.5revolutions to radians.
olga55 [171]

As we know that

1 revolution = 2\pi radian

now we know that

3.5 revolution = 3.5 rev\times \frac{2\pi rad}{1 rev}

3.5 rev = 21.98 rad

so it is 21.98 radian

8 0
3 years ago
Read 2 more answers
A string is stretched to a length of 308 cm and both ends are fixed. If the density of the string is 0.023 g/cm, and its tension
soldi70 [24.7K]

Answer:

Frquency=3,994Hz

Explanation:

Tension =967N

Density of string (μ)=0.023g/cm

Length of the stretched spring=308cm

Fundamental frequency for nth harmonic :

Fn=n/2L(√T/μ)

Substituting the given values to find the frequency :

f1=1/2(308cm) *(0.01m/1cm)[(√967N)/(0.023g/cm)(0.1kg)/(0.1kg/m)/(1g/cm)]

=6.16m[(√967N)/0.0023kg/m)]

=3,994.20Hz

Approximately,

The frequency will be =3,994Hz

7 0
3 years ago
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