Please help

Henry was giving the school toy for his birthday he can use it to tell how fast someone is moving at the moment in time the toy

pochemuha

Answer:

The toy must calculate the person's speed/velocity

Explanation:

Since the school toy given to Henry can be used to tell how fast someone is moving, the toy must be able to calculate the person's speed/velocity using the average distance covered by the person divided by time taken to cover the distance; average distance ÷ time taken.

The toy must be able to determine the parameters (average distance and time taken) in order to be able to calculate the person's speed/velocity

5 0

3 years ago

An airplane traveling at 201 m/s makes a turn. What is the smallest radius of the circular path (in km) that the pilot can make

egoroff_w [7]

Answer:

the smallest radius of the circular path is 8.1 km

Explanation:

The computation of the smallest radius of the circular path is given below:

Given that

V = Velocity = 201 m/s

a_c = acceleration = 5 m/s^2

radius = ?

As we know that

a_c = V^2 ÷ r

5 = 201^2 ÷ r

r = 201^2 ÷ 5

= 8,080.2 g

= 8.1 km

Hence, the smallest radius of the circular path is 8.1 km

4 0

2 years ago

A crane raises a crate with a mass of 150 kg to a height of 20 m. Given that

Virty [35]

Answer:

$\boxed {\boxed {\sf 29,400 \ Joules}}$

Explanation:

Gravitational potential energy is the energy an object possesses due to its position. It is the product of mass, height, and acceleration due to gravity.

$E_P= m \times g \times h$

The object has a mass of 150 kilograms and is raised to a height of 20 meters. Since this is on Earth, the acceleration due to gravity is 9.8 meters per square second.

m= 150 kg
g= 9.8 m/s²
h= 20 m

Substitute the values into the formula.

$E_p= 150 \ kg \times 9.8 \ m/s^2 \times 20 \ m$

Multiply the three numbers and their units together.

$E_p=1470 \ kg*m/s^2 \times 20 m$

$E_p=29400 \ kg*m^2/s^2$

Convert the units.

1 kilogram meter square per second squared (1 kg *m²/s²) is equal to 1 Joule (J). Our answer of 29,400 kg*m²/s² is equal to 29,400 Joules.

$E_p= 29,400 \ J$

The crate has 29,400 Joules of potential energy.

7 0

3 years ago

The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s

just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c) 78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0

3 years ago

Is velocity ratio of a machine affected by applying oil on it?Explain with reason.

disa [49]

Factors affecting friction

The intensity of friction depends on following factors: i) The area involved in friction. ii) The pressure applied on the surfaces. Force = Pressure ´ Area Frictional force will increase, if the area of contact will increase or if pressure applied on the surface increased.

Methods to reduce friction

i) Polish the contact surface. ii) Put oil or grease so that it fills in the small gaps of the flat parts. iii) Use ball bearings to reduce area of contact between rotating parts.

Lubrication

Following methods can be used to reduce friction: Oil is either thin or viscous. It depends upon SAE No. of oil. (SAE means Society of Automotive Engineers). If we use very viscous oil, it does not reach all the parts. Very thin oil will flows away easily and gets wasted. Grease is used in such cases. It is generally used around ball-bearing. Normal grease or oil is never used where there is high pressure, high temperature and high speed. Special lubricants are used in such cases. In cold season the oil becomes thick and in hot season it becomes thin. Therefore selection of lubrication also depends on the season. It is always advisable to refer operating manual of the equipment before selecting the lubricant.

6 0

3 years ago