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3 years ago

9

a 6-kg and a 4-kg ball are acted on by forces of equal size. if the large ball accelerates at 2 m/s2 what acceleration will the

small ball undergo

2 answers:

Dennis_Churaev [7]3 years ago

5 0

To answer this question, we need to find the force acting on the two objects.

f=ma
f=(6)(2)
f=12 newtons

So, now we can determine the acceleration of the second object using the same formula!

f=ma
12=(4)a
3=a

So the acceleration of the smaller object is 3 meters per second squared!

Anna007 [38]3 years ago

3 0

It would be 30 meters.

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Consider a stone in free fall on a planet with gravitational acceleration 3.4 m/s^2. Suppose you would like the stone to experie

Stella [2.4K]

Answer:

Angle of incline is 20.2978°

Explanation:

Given that;

Gravitational acceleration on a planet a = 3.4 m/s²

Gravitational acceleration on Earth g = 9.8 m/s²

Angle of incline = ∅

Mass of the stone = m

Force on the stone along the incline will be;

F = mgSin∅

F = ma

The stone has the same acceleration as that of the gravitational acceleration on the planet.

so

ma = mgSin∅

a = gSin∅

Sin∅ = a / g

we substitute

Sin∅ = (3.4 m/s²) / (9.8 m/s²)

Sin∅ = 0.3469

∅ = Sin⁻¹( 0.3469 )

∅ = 20.2978°

Therefore, Angle of incline is 20.2978°

8 0

3 years ago

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

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3 years ago

The minimum number of vectors of unequal magnitude required to

EleoNora [17]

Answer:

d

Explanation:

4 0

3 years ago

What is the speed of a truck traveling 10km in 10 minutes

user100 [1]

Answer: 58.8235 km/h

speed = distance/time

the distance is 10 km

the time is 10 minutes

the unit is not correct, so we first change minute to hour

so 10/60 is 0.166667, rounded to 0.17.

10 km/ 0.17 hours =

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1 year ago

a sphere of diameter 6•0cm is moulded into a thin uniform wire of diameter 0•2mm.calculate the length of the wire in metres

Scilla [17]

The length of the wire is 36 m.

<u>Explanation:</u>

Given, Diameter of sphere = 6 cm

We know that, radius can be found by taking the half in the diameter value. So,

$\text { sphere radius, } R=\frac{D}{2}=\frac{6}{2}=3 \mathrm{cm}=3 \times 10^{-2} \mathrm{m}$

Similarly,

$\text { wire radius, } r=\frac{0.2}{2}=0.1 \mathrm{mm}=1 \times 10^{-3} \mathrm{m}$

We know the below formulas,

$\text {volume of sphere}=\frac{4}{3} \times \pi \times R^{3}$

$\text {volume of wire}=\pi \times r^{2} \times l$

When equating both the equations, we can find length of wire as below, where $\pi=\frac{22}{7}$

$\frac{4}{3} \times \pi \times R^{3}=\pi \times r^{2} \times l$

$\frac{4}{3} \times \frac{22}{7} \times\left(3 \times 10^{-2}\right)^{3}=\frac{22}{7} \times\left(1 \times 10^{-3}\right)^{2} \times l$

The $\pi$ value gets cancelled as common on both sides, we get

$\frac{4}{3} \times 27 \times 10^{-6}=10^{-6} \times l$

The $10^{-6}$ value gets cancelled as common on both sides, we get

$l=4 \times 9=36 m$

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3 years ago