All categories

2 years ago

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

1 answer:

5 0

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

You might be interested in

You may have noticed runaway truck lanes while driving in the mountains. These gravel-filled lanes are designed to stop trucks t

kati45 [8]

Answer:

The coefficient of kinetic friction $\mu_k = 0.724$

Explanation:

From the question we are told that

The length of the lane is $l = 36.0 \ m$

The speed of the truck is $v = 22.6\ m/s$

Generally from the work-energy theorem we have that

$\Delta KE = N * \mu_k * l$

Here N is the normal force acting on the truck which is mathematically represented as

$\Delta KE$ is the change in kinetic energy which is mathematically represented as

$\Delta KE = \frac{1}{2} * m * v^2$

=> $\Delta KE = 0.5 * m * 22.6^2$

=> $\Delta KE = 255.38m$

$255.38m = m * 9.8 * \mu_k * 36.0$

=> $255.38 = 352.8 * \mu_k$

=> $\mu_k = 0.724$

6 0

3 years ago

Suppose you lived in a pre-industrial society and needed to lift a heavy (20 kg) block a height of 5 m and had two choices for h

igomit [66]

Let's break the question into two parts:

1) The force needed in Ramp scenario.
2) The effort force needed in the lever scenario.

1. Ramp Scenario:
In an incline, the only component of cart's weight(mg) that is in the direction of motion is $mgsin \alpha$ . Therefore the effort force in this case must be equal or greater than $mgsin \alpha$ .

Now we need to find $\alpha$ . $\alpha$ is the angle between the incline of the ramp and the ground.

Since the height is 5m and the length of the ramp is 8m, $sin \alpha$ would be 5/8 or 0.625. Now that you have $sin \alpha$ , mutiple it with mg.

=> m*g* $sin \alpha$ = 20 * 10 * 5 / 8. (Taking g = 10 m/s² for simplicity) = 125N
Therefore, the minimum Effort force you would require in this case is 125N.

2. Lever Scenario:
Just apply "moment action" in this case, which is:
$F_{e} d_{e} = F_{r} d_{r}$

$F_{e}$ = ?

$F_{r}$ = mg = 20 * 10 = 200N
$d_{e}$ = 10m
$d_{r}$ = 1m

Plug-in the values in the above equation:
$F_{e}$ = 200/10= 20N

As 20N << 125N, the best choice is to use lever.

4 0

3 years ago

A compunds whose empirical formula is xf3 consists of 65% f by mass. what is the atomic mass of x

Reika [66]

Empirical formula of compound is XF3

Compound consist of 65% F

In 100g of compound there is 65 g of F

= 65 / 19 moles of Fluorine = 3.421 moles

So moles of X = 3.421 / 3 = 1.140 moles

And in 100 g X consist of 35 g

So the molar mass of X = 35 / 1.140 = 30.71 g = 31 approximately

And it is the mass of phosphorus

So the empirical formula for the compound is PX3

4 0

3 years ago

Read 2 more answers

Our solar system is located very close to the center of the Milky Way galaxy, outside of the spiraling arms.

inna [77]

FALSE
IF IT WAS THEN IT WOULd BE TRUE

4 0

3 years ago

Read 2 more answers

At a sports car rally, a car starting from rest accelerates uniformly at a rate of 5 m/s/s over a straight-line distance of 291

Bogdan [553]

Answer:

10.8 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Acceleration (a) = 5 m/s/s

Distance travelled (s) = 291 m

Time (t) taken =?

We can calculate the time taken for the car to cover the distance as follow:

s = ut + ½at²

291 = 0 × t + ½ × 5 × t²

291 = 0 + 2.5 × t²

291 = 2.5 × t²

Divide both side by 2.5

t² = 291 / 2.5

t² = 116.4

Take the square root of both side

t = √116.4

t = 10.8 s

Thus, it will take the car 10.8 s to cover the distance.

8 0

2 years ago