All categories

3 years ago

Can someone help me with PE questions

Physics

2 answers:

skelet666 [1.2K]3 years ago

6 0

1. What does PE stand for? Physical Education

2. Name one common PE exercise? Your choice A knee push up B piano keys

9966 [12]3 years ago

5 0

Not until we know what the questions are.

You might be interested in

A 970-kg sports car collides into the rear end of a 2300-kg SUV stopped at a red light. The bumpers lock, the brakes are locked,

vlabodo [156]

Answer:

22.73 m/s or 81.72 kph

Explanation

We can find the combined mass of both cars as

970 kg + 2300 kg = 3270 kg.

Then the normal force of the cars can be calculated as

F(n)= mg

Where g is acceleration due to gravity 9.8m/s^2

3270 kg ×9.8 = 32046 kg*m/s^2.

coefficient of kinetic friction between tires and road to be 0.80 × F(n)

Then the frictional force can be calculated as

= (32046kg*m/s^2 × 0.80 )

= 25636.8 kg*m/s^2

We can now calculate the work done that was used stopping the cars as

Frictional force × distance

(25636.8 kg*m/s^2 ) × 2.9m= 74346.72kg*m^2/s^2

From kinetic energy formula, the combined velocity of the car can be determined

E=0.5 M V²

√(2E/M) = V

√(2*74346.72kg*m^2/s^2 / 3270 kg) = V

V= √ (45.472)

V=6.743293m/s

the momentum of both cars can be determined as

6.743293m/s * 3270 kg

= 22050.57kg*m/s

Now the final momentum of both cars must be equal to the the momentum of

the sports car just prior to the collision. Therefore, the speed of the sports car at impact.

=(22050.57 kg*m/s) / 970 kg = 22.73 m/s

We can convert that to km/h.

22.73 m/s * 3600 s/h / 1000 m/km = 81.72 kph

7 0

2 years ago

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). if a particular disk is spun at 79

NeX [460]

(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²

4 0

3 years ago

You are part of the designing team for a new solar thermal power plant. The plant will use thermal energy storage to produce ele

Harman [31]

Explanation:

Below is an attachment containing the solution.

4 0

3 years ago

What gas law applies to the situation you have described above. Why?

Allushta [10]

If you give us the situation described then I'll be able to help.

6 0

3 years ago

An object is hung from a spring balance attached to the ceiling of an elevator cab. The balance reads 65 N when the cab is stand

lana66690 [7]

Answer:

(A) Reading will be 65 N

(B) Net force on the elevator will be 49.076 N

Explanation:

We have given the balance force = 65 N

Acceleration due to gravity $g=9.8m/sec^2$

We know that W=mg

So $65=m\times 9.8$

m = 6.632 kg

(a) In first case as the as the speed is constant so the force on the elevator will be 65 N

(B) In second case as the elevator is decelerating at a rate of $2.4m/sec^2$

So net acceleration = 9.8-2.4= $7.4m/sec^2$

So net force on elevator will be = m× net acceleration = 6.632×7.4 = 49.076 N

7 0

3 years ago