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3 years ago

A car drives on a highway with a speed of 68mi/hr. What is the speed in km/hr?

Physics

1 answer:

spin [16.1K]3 years ago

3 0

Answer:

Speed in km/hr will be 109.412 km/hr

Explanation:

We have given speed of the car on a highway = 68 mi/hr

We have to find the speed in km/hr

For this first we have to change mi to km

We know that 1 mile = 1.609 km

Speed is the ratio of distance and time

So 68 mi/hr $=68\times 1.609km/hr=109.412km/hr$

So the speed in km/hr will be 109.412 km /hr

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1.Explain frame of reference. How is it different when you are riding inside a car or standing by the highway?

galben [10]

Answer:

See explanation

Explanation:

A frame of reference simply refers to a set of coordinates which is used to determine the position and velocity of objects all the objects found in that particular frame. The person inside the car and standing by the highway are in different frames of reference.

Newtons first law of motion states that an object will continue to be at rest or in a state of uniform velocity until it is acted upon by an external force.

Newtons second law states that rate of change of momentum is directly proportional to the impressed force.

Newtons third law states that action and reaction are equal and opposite.

The action of applying brakes leads to an equal reaction of a sudden forward movement. Seat belts help to prevent the occupants of a car from being injured by the sudden stoppage of the car.

Force is simply defined as a push or pull.

Balanced forces do not lead to acceleration of a body but an unbalanced for leads to acceleration of a body.

strong nuclear force, electromagnetic force, weak nuclear force and gravity

Power is the rate of doing work while week is said to be done when the force applied moves a distance in the direction of the force. Power is defined as work/time.

8 0

3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

3 years ago

A swimmer is swimming to the left with a speed of 1.0 m/s when she starts to speed up with constant acceleration. The swimmer re

kolezko [41]

Answer:

Correct answer: t = 2.86 seconds

Explanation:

We first use this formula

V² - V₀² = 2 a d

where V is the final velocity (speed), V₀ the initial velocity (speed),

a the acceleration and d the distance.

We will calculate the acceleration from this formula

a = (V² - V₀²) / (2 d) = (2.5² - 1²) / (2 · 5) = (6.25 - 1) / 10 = 5.25 / 10

a = 0.525 m/s²

then we use this formula

V = V₀ + a t => t = (V - V₀) / a = (2.5 - 1) / 0.525 = 1.5 / 0.525 = 2.86 seconds

t = 2.86 seconds

God is with you!!!

6 0

3 years ago

The particle, initially at rest, is acted upon only by the electric force and moves from point a to point b along the x axis, in

bezimeni [28]

1) Potential difference: 1 V

2) $V_b-V_a = -1 V$

Explanation:

When a charge moves in an electric field, its electric potential energy is entirely converted into kinetic energy; this change in electric potential energy is given by

$\Delta U=q\Delta V$

where

q is the charge's magnitude

$\Delta V$ is the potential difference between the initial and final position

In this problem, we have:

$q=4.80\cdot 10^{-19}C$ is the magnitude of the charge

$\Delta U = 4.80\cdot 10^{-19}J$ is the change in kinetic energy of the particle

Therefore, the potential difference (in magnitude) is

$\Delta V=\frac{\Delta U}{q}=\frac{4.80\cdot 10^{-19}}{4.80\cdot 10^{-19}}=1 V$

Here we have to evaluate the direction of motion of the particle.

We have the following informations:

- The electric potential increases in the +x direction

- The particle is positively charged and moves from point a to b

Since the particle is positively charged, it means that it is moving from higher potential to lower potential (because a positive charge follows the direction of the electric field, so it moves away from the source of the field)

This means that the final position b of the charge is at lower potential than the initial position a; therefore, the potential difference must be negative:

$V_b-V_a = - 1V$