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3 years ago

what is the average acceleration of a car if a car rest from 0m/s and reaches a final velocity is 50m/s in 10m/s

Physics

1 answer:

GuDViN [60]3 years ago

6 0

Answer:

Assuming that you meant the final velocity of 50 m/s was reached in 10 s, the answer would be 5 m/s^2.

Explanation:

$V_{f} = V_{i} + at$

So we update that with the values that we have.

$50 = 0 +a(10)$

then simplify that using algebra to solve for a and we get 5 m/s^2

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Now in "real life," this automobile is cruising at 20.5 m/s (equal to 73.8 km/hr) when it is about to hit a pedestrian stuck in

algol13

Answer:

He needs 1.53 seconds to stop the car.

Explanation:

Let the mass of the car is 1500 kg

Speed of the car, v = 20.5 m/s

He will not push the car with a force greater than, $F=2\times 10^4\ N$

The impulse delivered to the object is given by the change in momentum as :

$F\times t=mv\\\\t=\dfrac{mv}{F}\\\\t=\dfrac{1500\times 20.5}{2\times 10^4}\\\\t=1.53\ s$

So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.

5 0

3 years ago

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Romashka [77]

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

6 0

3 years ago

You use 350 W of power to move a 7.0 N object 5 m.<br> How long did it take?

PilotLPTM [1.2K]

Answer:

0.1 second

Explanation:

We are given;

Power; P = 350 W

Force; F = 7 N

Distance; d = 5 m

Formula for power is;

P = workdone/time taken

Workdone = F × d

Thus;

350 = (7 × 5)/t

t = 35/350

t = 0.1 second

4 0

3 years ago

A 11,000-kg train car moving due east at 21.0 m/s collides with and couples to a 23,000-kg train car that is initially at rest.

Tomtit [17]

Answer:

v = 6.79 m/s

Explanation:

It is given that,

Mass of a train car, m₁ = 11000 kg

Speed of train car, u₁ = 21 m/s

Mass of other train car, m₂ = 23000 kg

Initially, the other train car is at rest, u₂ = 0

It is a case based on inelastic collision as both car couples each other after the collision. The law of conservation of momentum satisied here. So,

$m_1u_1+m_2u_2=(m_1+m_2)V$

V is the common velocity after the collisions

$V=\dfrac{m_1u_1}{(m_1+m_2)}\\\\V=\dfrac{11000\times 21}{(11000+23000)}\\\\V=6.79\ m/s$

So, the two car train will move with a common velocity of 6.79 m/s.

6 0

3 years ago

The pressure in a compressed air storage tank is 1200 kPa. What is the tank’s pressure in (a) kN and m units; (b) kg, m, and s u

elixir [45]

Explanation:

Given data

Pressure P=1200 kPa

To find

Pressure in

(a) kN/m²

(b) kg/m.s²

Solution

For Part (a)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )\\P=1200kN/m^{2}$

For Part (b)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )\\P=1,200,000kg/m.s^{2}$

For Part (c)

$P=(1200kPa)(\frac{1kN/m^{2} }{1kPa} )(\frac{1000kg.m/s^{2} }{1kN} )(\frac{100m}{1km} )\\P=1,200,000,000kg/km.s^{2}$

8 0

4 years ago