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vodka [1.7K]
3 years ago
9

An initially motionless test car is accelerated uniformly to 145 km / h 145 km/h in 8.78 s 8.78 s before striking a simulated de

er. The car is in contact with the faux fawn for 0.815 s , 0.815 s, after which the car is measured to be traveling at 76.5 km / h . 76.5 km/h. What is the magnitude of the acceleration of the car before the collision?
Physics
1 answer:
vfiekz [6]3 years ago
3 0

Answer:

a = 4.59 m/s²

Explanation:

given,

initial speed of the car = 0 m/s

final speed of the car = 145 km/h

1 Km/h = 0.278 m/s

145 Km/h = 145 x 0.278 = 40.31 m/s

time, t = 8.78 m/s

Acceleration of car before collision

a = \dfrac{\Delta v}{t}

a = \dfrac{v_f- v_i}{t}

a = \dfrac{40.31 - 0 }{8.78}

a = 4.59 m/s²

The magnitude of acceleration before collision is 4.59 m/s²

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What is one way you are using nonrenewable resources?
lutik1710 [3]

Putting gas in your car is one way you are using non-renewable resources.

You could Reduce, Recycle and Re-use items. You could manage what you do with you're water. Reducing and reusing products cuts down on manufacturing pollution, just as the use of recycled instead of virgin materials prevents pollution in industrial processes. You could add food coloring to your toilet tank. You could also do this: use sprinkler heads that sprinkle mist out instead of droplets of water.  to avoid losing water to evaporation.

6 0
4 years ago
6. A warehouse employee is pushing a 15.0 kg desk across a floor at a constant speed of 0.50 m/s. How much work must the employe
MariettaO [177]

Answer:

7.5 J

Explanation:

To answer the question given above, we need to determine the energy that will bring about the speed of 1 m/s. This can be obtained as follow:

Mass (m) = 15 Kg

Velocity (v) = 1 m/s

Energy (E) =?

E = ½mv²

E = ½ × 15 × 1²

E = ½ × 15 × 1

E = ½ × 15

E = 7.5 J

Therefore, to change the speed to 1 m/s, the employee must do a work of 7.5 J.

3 0
3 years ago
a 60 kg woman in a elevator is accelerating upward at a rate of 1.2 m/s2. What is the gravitational force acting upon the woman?
Flura [38]

The gravitational force acting upon the woman is equal to <u>-588.6N</u>

Why?

To solve the problem, we need to consider that two forces are acting upon the woman, the first one is related to her weight and the second one is related to the acceleration of the elevator.

Gravitational force acting upon the woman:

Force=m*gravityacceleration\\\\Force=60kg*-9.81\frac{m}{s^{2}}=-588.6N

Hence, we have that the gravitational force acting upon the woman is equal to -588.6N.

Have a nice day!

5 0
3 years ago
A car travels due east with a speed of 38.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth
DiKsa [7]

Answer: 116.926 km/h

Explanation:

To solve this we need to analise the relation between the car and the Raindrops. The cars moves on the horizontal plane with a constant velocity.

Car's Velocity (Vc) = 38 km/h

The rain is falling perpedincular to the horizontal on the Y-axis. We dont know the velocity.

However, the rain's traces on the side windows makes an angle of 72.0° degrees. ∅ = 72°

There is a relation between this angle and the two velocities. If the car was on rest, we will see that the angle is equal to 90° because the rain is falling perpendicular. In the other end, a static object next to a moving car shows a horizontal trace, so we can use a trigonometric relation on this case.

The following equation can be use to relate the angle and the two vectors.

Tangent (∅) = Opposite (o) / adjacent (a)

Where the Opposite will be the Rain's Vector that define its velocity and the adjacent will be the Car's Velocity Vector.

Tan(72°) = Rain's Velocity / Car's Velocity

We can searching for the Rain's Velocity

Tan(72°) * Vc = Rain's Velocity

Rain's Velocity = 116.926 km/h

3 0
4 years ago
Need help on physics/momentum ​
koban [17]

Answer:

vf=11.67m/s

Explanation:

vf:

Δp=m(vf-vi)

80=12(vf-5)

80=12vf-60

140=12vf

vf=11.67m/s

pf:

p=m*v

p=12*11.67

p=140.04 N*s

Hope this helps

5 0
3 years ago
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