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3 years ago

What kind of energy does an unlit match have?

Physics

1 answer:

Rudiy273 years ago

6 0

It’s is none of the above answers.
It is d. Potential energy

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A 2800 kg speedboat starting from rest attains a speed of 16 m/s in 8.0 s as a combination of 1200 N of air resistance and water

denis23 [38]

Answer:

For me

Explanation:

power=workdone/time

power= 128×1200/8

power=19200

6 0

3 years ago

Physics 1 Course. please help.

Aleks [24]

Answer:

a. 1.64 m/s²

Explanation:

Centripetal acceleration is the square of tangential velocity divided by the radius.

a = v²/r

First, convert km/h to m/s.

30.0 km/h (1000 m/km) (1 h / 3600 s) = 8.33 m/s

Find the acceleration.

a = (8.33 m/s)² / (42.4 m)

a = 1.64 m/s²

6 0

3 years ago

A horizontal force of 50 N is applied to the object.

USPshnik [31]

Answer:

Mass of object (m) = 5.102 kg

Explanation:

Given:

Horizontal Force (F) = 50 N

Find:

Mass of object (m) = ?

Computation:

We know that, acceleration due to gravity (g) = 9.8 m/s²

⇒ Horizontal Force (F) = mg

⇒ 50 N = m (9.8 m/s²)

⇒ Mass of object (m) = 50 / 9.8

⇒ Mass of object (m) = 5.102 kg

Mass of object (m) is 5.1 kg (Approx)

3 0

3 years ago

When a bicycle coasts uphill, it moves slower and slower as it climbs. Why?

astra-53 [7]

Letter B because it is gaining more potential energy as it SLOWLY climbs up the hill.

the less motion the more potential energy there is

6 0

3 years ago

Read 2 more answers

A small mailbag is released from a helicopter that is descending steadily at 3 m/s.

mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Package is 44.1 meter below helicopter

<u>Explanation:</u>

a) We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

Initial velocity = 3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = 3+9.8*3 = 32.4 m/s

Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m$

Distance traveled by package = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 $m/s^2$ and time = 3 seconds.

v = -3+9.8*3 = 26.4 m/s

Speed of mailbag after 3 seconds = 26.4 m/s

Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 $m/s^2$ .

$s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m$

Distance traveled by helicopter = 9 meter.

Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 $m/s^2$ .

$s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m$

Distance traveled by package = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0

3 years ago