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natita [175]
3 years ago
11

Determine the change in velocity of a car that starts at rest and has a final velocity of 20 miles per second North

Physics
1 answer:
Katen [24]3 years ago
5 0

Answer: The change in velocity is 20mph

Explanation: The change in velocity is the difference between the final velocity and the initial velocity.

The initial velocity is 0 and the final velocity is 20mph.

Using the formula dV=Vf-Vi

dV=20-0

dV=20mph North

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When gases, liquids, or solids are in contact with a moving object, the flow of _____ occurs due to frictional forces
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When gases, fluids, or other solids are in contact with a moving object

heat is produced due to friction.

7 0
3 years ago
Read 2 more answers
You toss a rock up vertically at an initial speed of 39 feet per second and release it at an initial height of 6 feet. The rock
3241004551 [841]

Answer:

2.583 s, 29.77 ft and 1.219 s

Explanation:

Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground

H = 6 ft, the height from which it is tossed

V₁ = V + gt = V - gt

at maximum height the body came to rest momentarily V₁ = 0

0 = V - gt

-V = -gt

- 39 / -32 = t

t time to reach maximum height = 1.219 s

To Maximum height reached can be calculated with the formula

V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock

where V₁ is the final velocity at maximum height which = 0

0 = V² - 2g(y-H) where y is the distance traveled from the ground

-V² = -2g(y-H)

₋V² / -2g = y-H

(V²/2g) + H = y in ft

(39² / (2 × 32)) + 6

y = 29.77 ft

The total time it will be in air can be calculated with the formula below

y = H + Vt - 0.5gt² from y-H = ut + 0.5at²

0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)

0.5gt² - Vt - H = 0

using quadratic formula

- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)

(V ± √ (V² + 2gH)) ÷ g

substitute the values into the expression

t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32

t = (39 + √(1521 +384))/32 = (39 + √1905) / 32  = 2.583 s

t = (39 - √1905) / 32 =  -0.15 s

The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds

8 0
2 years ago
Driving Zone 7 corresponds with
stepan [7]
Your driving zone refers to the areas of space around your car, it refers to all the area around your car as far as your eyes can see. 
Each car has seven zones numbered from 1 to 7. Driving zone 7 corresponds with THE SPACE YOUR VEHICLE IS OCCUPYING. The other zones are as follows:
zone 1 = area directly infront of your car
zone 2 = your left lane
zone 3 = your right lane
zone 4 = left rear of your car
zone 5 = right rear of your car 
zone 6 = area directly behind your car.
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3 0
2 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
If potential energy at point A is 12 joules, what is the potential energy at D?
Irina-Kira [14]

Answer:

In D: 3J

Explanation:

Potential energy: Ep=mgh where m is the mass, h altitude.

In point A: h=20cm=0.2m

Epa=12=0.2×mg. Thus mg=12/0.2=60N

For point D: hd=5cm=0.05m

Epd=mg×0.05=60×0.05=3J

5 0
3 years ago
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