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makvit [3.9K]
2 years ago
15

A child tugs on a rope attached to a 0.62 kg toy with a horizontal force of 16.3 N. A puppy pulls the toy in the opposite direct

ion with a force of 15.8 N. What is the acceleration of the toy
Physics
1 answer:
sattari [20]2 years ago
5 0

Answer:

a=51.77\ m/s^2

Explanation:

Given that,

The mass of a toy, m = 0.62 kg

Force with which a child pull the toy = 16.3 N

The force with which the toy pulled in the opposite direction = -15.8 N

We need to find the acceleration of the toy. Let F be the net force acting on the toy. It is equal to :

F = 16.3 N - (-15.8 N)

= 32.1 N

Let a be the acceleration of the toy. Using Newton' second law of motion to find it.

F = ma

a=\dfrac{F}{m}\\\\a=\dfrac{32.1}{0.62}\\\\a=51.77\ m/s^2

So, the acceleration of the toy is 51.77\ m/s^2.

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What is the radius of an automobile tire that turns with a period of 0.091 s and has a linear
faust18 [17]

Answer:

1) The radius of the tire is approximately 0.28966 meters

2) The centripetal force is the force that keeps a body moving on a circular path

Explanation:

1) The linear speed of the automobile tire = 20.0 m/s

The period with which the tire turns = 0.091 s

The period = The time it takes to make a complete turn

Therefore;

The number of turns in 1 second = 1/0.091 ≈ 10.989 turns

The distance covered with 10.989 turns, assuming no friction = 10.989 × The circumference of the tire

∴ The distance covered with 10.989 turns, assuming no friction = 10.989 × 2 × π × Radius of the tire

From the speed of the car, 20.0 m/s, we have;

The distance covered in 1 second = 20.0 meters

Therefore;

10.989 × 2 × π × Radius of the tire = 20.0 meters

Radius of the tire = (20.0 meters)/(10.289 × 2 × π) ≈ 0.28966 meters

The radius of the tire ≈ 0.28966 meters

2) The centripetal force is the force required to maintain the curved motion of an object, and having a direction towards the center of the rotary motion.

The centripetal force is given by the formula, F = \dfrac{m \cdot v^2}{r}

Where;

F = The centripetal force

m = The mass of the object

v = The linear velocity of the object

r = The radius of the rotational motion.

3 0
2 years ago
A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
2 years ago
Use the formula a=h/n to find the batting average a of a batter who has h hits in n times at bat. Solve the formula for h. if a
Inessa [10]
A = h / n => h = a*n

a = 0.290 hit / time
n = 300 times

=> h = 0.290 hit / time * 300 time = 87 hits

Answer: 87 hits
4 0
2 years ago
A 0.200 kg plastic ball moves with a velocity of 0.30 m/s. It collides with a second plastic ball of mass 0.100 kg, which is mov
zzz [600]

Answer:

0.22m/s

Explanation:

The total momentum of the System is conserved. Total momentum of the system before the collision is equal to the total momentum of the system after collision. The total momentum is the sum of individual momentum of all the objects in that system.

momentum of an object = mass* velocity

Total Momentum before collision = 0.2*0.3 + 0.1*0.1= 0.07 kg⋅m/s;

Total momentum after collision = 0.1*0.26 + 0.2*x = 0.07;

Solve for x.

4 0
2 years ago
If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level loc
geniusboy [140]

Complete question:

The classic Goodyear blimp is essentially a helium balloon— a big one, containing 5700 m³ of helium. If the envelope and gondola have a total mass of 4300 kg, what is the maximum cargo load when the blimp flies at a sea-level location? Assume an air temperature of 20°C.

Answer:

52.4 kN

Explanation:

The helium at 20°C has a density of 0.183 kg/m³, and the cargo load is the weight of the system, which consists of the envelope, the gondola, and the helium.

The helium mass is the volume multiplied by the density, thus:

mHe = 5700 * 0.183 = 1043.1 kg

The total mass is then 5343.1 kg. The weight is the mass multiplied by the gravity acceleration (9.8 m/s²), so:

W = 5343.1*9.8

W = 53362.38 N

W = 52.4 kN

5 0
2 years ago
Read 2 more answers
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