All categories

3 years ago

what is the average speed of a car that travels 40mph for 1 hour and 60 mph in another hour PLEASE HELP WILL MARK BRAINLIEST

Physics

1 answer:

suter [353]3 years ago

8 0

Average speed = (distance covered) / (time to cover it)

-- Traveling at 40 mph for 1 hour, the distance covered is 40 miles.

-- Traveling at 60 mph for 1 hour, the distance covered is 60 miles.

-- Total distance covered = (40 miles) + (60 miles) = 100 miles

-- Total time = (1 hour) + (1 hour) = 2 hours

-- Average speed = (100 miles) / (2 hours)

<em>Average speed = 50 miles per hour</em>

You might be interested in

If a an object is traveling at a rate of 5 meters per second squared with a force of 10 Newtons, what is the mass of the object?

Svetllana [295]

Answer:

2 kg

Explanation:

Acceleration = 5 m/s^2

Force = 10 N

Force = mass * acceleration

mass = force / acceleration

mass = 10 / 5

mass = 2 kg

8 0

3 years ago

When you must add three vectors together, what is not true this process? You must only give a magnitude of the resultant vector

vovangra [49]

The addition of any numbers of vector provide the magnitude as well as the direction of the resultant vector, hence the mentioned first option is not true.

The addition of vector required to connect the head of the one vector with the tail of the other vector and any vector can be moved in the plane parallet to the previous location, so, the mentioned second and third options are true.

4 0

3 years ago

A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?

rosijanka [135]

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

5 0

3 years ago

Each driver has mass 79.0 kg. Including the masses of the drivers, the total masses of the vehicles are 800 kg for the car and 4

Mademuasel [1]

Answer:

Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

3 0

3 years ago

At which point does the planet have the least gravitational force acting on it?

Elza [17]

Answer:

At which point does the planet have the least gravitational force acting on it?

Explanation:

In an elliptical orbit, when a planet is at its furthest point from the Sun, it is under the least amount of gravity, meaning that the force of gravity is strongest when it is closest.

5 0

3 years ago

Read 2 more answers