Ask question

Ask question

All categories

4 years ago

12

A certain unbalanced force gives a 5kg object an acceleration of 15 m/s2. What would be the acceleration if the same force was a

pplied to 30 kg object ?

1 answer:

stira [4]4 years ago

7 0

Answer:

a2 = 2.5 m/s2

Explanation:

F1 = m1 a1 We use the same force so F1 = F2

= 5kg × 15m/s2 F2 = m2 a2

= 75N a2 is required

a2 = F2 / m2

= 75N / 30 kg

= 2.5 m/s2

You might be interested in

Which best describes what happens to sunlight after passing through the keyhole of a door

Law Incorporation [45]

Given the answers I would say B, Is your best choice because when the light goes through a hole it spreads out, because most light rays (<) outward.

7 0

4 years ago

Which of the following is Ohm's Law?<br> I = V/R<br> V = I/R<br> R = I x R<br> V = R/I

Digiron [165]

V=I*R is the correct answer

5 0

3 years ago

Read 2 more answers

A small remote-control car with a mass of 1.65 kg moves at a constant speed of v = 12.0 m/s in a vertical circle inside a hollow

Reil [10]

Answer:

$N=63.69N$

Explanation:

The normal force exerted on the car by the walls of the cylinder at the bottom of the vertical circle will be such that when substracted to the weight it must give the centripetal force, since at that point on the vertical $F_{cp}=N-W=N-mg$

We also know that the equation for the centripetal force is:

$F_{cp}=ma_{cp}=\frac{mv^2}{r}$

Mixing both equations we get:

$N-mg=\frac{mv^2}{r}$

$N=mg+\frac{mv^2}{r}$

Which for our values means:

$N=(1.65Kg)(9.8m/s^2)+\frac{(1.65Kg)(12m/s)^2}{(5m)}=63.69N$

8 0

3 years ago

A Cessna aircraft has a liftoff speed of 120 km/h What minimum constant acceleration does the aircraft require to be airborne af

Lady_Fox [76]

Answer:

<h3>2.3125m/s²</h3>

Explanation:

Using the equation of motion v² = u²+2aS

v is the final velocity = 120km/hr

120km/hr = 120 * 1000/1 * 3600 = 33.3m/s

u is the initial velocity = 0m/s

a is the acceleration

S is the distance covered = 240m

On substituting the given parameters

33.3² = 0²+2a(240)

33.3² = 480a

1110 = 480a

a = 1110/480

a = 2.3125m/s²

Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²

4 0

4 years ago

The metal cover that seals the top of the cylinder

padilas [110]

Answer:a lid

Explanation bob pulled the lid off the jar of pickles

5 0

3 years ago

Read 2 more answers