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svlad2 [7]
4 years ago
5

Please help. Brainliest will be given! 25 points. Show all work.

Physics
1 answer:
Drupady [299]4 years ago
4 0

Explanation:

Given:

v₀ₓ = 15 m/s cos 20° = 14.10 m/s

aₓ = 0 m/s²

v₀ᵧ = 15 m/s sin 20° = 5.13 m/s

aᵧ = -9.8 m/s²

t = 1.5 s

Find: Δx and Δy

Δx = v₀ₓ t + ½ aₓ t²

Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²

Δx = 21.1 m

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²

Δy = -3.33 m

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An nfl linebacker can go from 0 m/s to 2.5 m/s in 2.5s. What is his acceleration?
dybincka [34]

Answer:

a= 1 m/s^2

Explanation:

a= (Vf-Vi)/t

a=(2.5m/s - 0m/s)/2.5 s

a=2.5 m/s / 2.5 s

a= 1 m/s^2

5 0
3 years ago
(c) Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What f
ziro4ka [17]

Answer:

Force, F = 2356.19 N

Explanation:

Given that,

Diameter of the circular tip, d = 1 mm = 0.001 m

Its radius, r = 0.0005 m

Pressure created, P=3\times 10^9\ N/m^2

To find,

Force exerted on the nail.

Solution,

The pressure exerted by an object is equal to the force acting per unit area. Its formula is given by :

P=\dfrac{F}{A}

F=P\times A

F=3\times 10^9\ N/m^2\times \pi (0.0005\ m)^2

F = 2356.19 N

Therefore, the force exerted on the nail is 2356.19 N.

6 0
3 years ago
The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure (in atm) du
frozen [14]

Answer:

P = 103867260 atm

Explanation:

The pressure at the bottom of any liquid column is equal to product of density of the liquid , gravitational acceleration constant (g) and height of the water column

Thus, P = \rho*g*h

Substituting the given values, we get -

P = 1029 kg/m3 * 9.8 m/s^2  *10.3 *1000 meters

P = 103867260 atm

8 0
3 years ago
Electron can only lose energy by transition from one allowed orbit to another ______ electromagnetic radiation
ozzi

Explanation:

Please solve this question as soon as possible. I need help with a solution

5 0
3 years ago
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attach
Lisa [10]

Answer:

<em>The time interval required to lift the spacecraft to this specified height is 123.94 seconds</em>

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = <em>14691.456 J  </em> is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74<em> J</em>

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

P = \frac{w}{t}

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = <em>123.94 seconds</em>

8 0
4 years ago
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