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3 years ago

If the same strength force is exerted on two objects why might they be affected differently?

Physics

1 answer:

Helen [10]3 years ago

6 0

Answer:

Due to their mass

Explanation:

If the same strength force is exerted on two objects they might be affected differently due to their mass.

The way objects reacts to a force impact is largely dependent on their mass.

The mass of an object is the amount of matter it contains.

According to the Newton's second law of motion, the force applied on a body is directly proportional to the product of its mass and acceleration.

Force = mass x acceleration

Therefore, the way a force acts on a body is a function of its mass.

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In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 32.0 mi/gal , how many liters of gasoline would

luda_lava [24]

Answer:

10.44 liters of gasoline.

Explanation:

First, we need to convert the units of the car from mi/gal to km/L as follows:

$mileage = 32.0 \frac{mi}{gal}*\frac{1.609 km}{1 mi}*\frac{1 gal}{3.785 L} = 13.60 km/L$

That means that for every liter of gasoline the car travels 13.60 kilometers.

So, to complete a 142-km trip in Europe the volume of gasoline needed is:

$V = \frac{1 L}{13.60 km}*142 km = 10.44 L$

Therefore, to complete a 142-km trip in Europe we need to buy 10.44 liters of gasoline.

I hope it helps you!

4 0

3 years ago

You construct a circuit containing some component C, along with other circuit elements. You want to simultaneously measure the c

Blababa [14]

Answer:

The position of the ammeter is in series down the component. While the voltmeter must be connected in parallel. Its position is the gap through the component C.

The positions of one or the other are indicated in the image.

Explanation:

An ammeter is defined as a device used to measure current. Its unit is the ampere. While a voltmeter is used in the measurement of the potential difference between two points. Its unit is the volt. The ammeter must be connected in series with the point at which the current is to be measured, while the voltmeter must be connected in parallel.

3 0

3 years ago

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block

borishaifa [10]

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

$T = 2\pi \sqrt{\frac{m}{k} }$

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

$T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\$

Now, determine the amplitude of oscillation, A;

$E = \frac{1}{2} kA^2$

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

$E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm$

Therefore, the amplitude of the oscillation is 2.82 cm

8 0

3 years ago

A cable that weighs 8 lb/ft is used to lift 900 lb of coal up a mine shaft 650 ft deep. Find the work done. Show how to approxim

Harrizon [31]

Answer:

2275000 lb.ft

Explanation:

Let work done on the cable be denoted by: W_ca

Let work done on the coal be denoted by: W_co

Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.

Meanwhile let δx be the length of the segment.

We are told the cable weighs 8 lb/ft. Thus;

Work done on one segment = 8 × δx × x = 8x•δx

Therefore, work done on cable is;

W_ca = ∫8x•δx between the boundaries of 0 and 650

Thus;

W_ca = 4x² between the boundaries of 0 and 650

W_ca = 4(650²) - 4(0²)

W_ca = 1,690,000 lb.ft

Workdone on the 900 lb of coal will be calculated as;

W_co = 900 × 650

W_co = 585000 lb.ft

Thus,

Total work done = W_ca + W_co

Total workdone = 1690000 + 585000

Total workdone = 2275000 lb.ft

6 0

3 years ago

How much energy is required to change a 39.0-g ice cube from ice at -14.0°c to steam at 108°c?

Ann [662]

Heat is added to the mass as follows:
Q_-14-0 = mC_iΔT = 39*2.06*(0--14) = 39*2.06*14 = 1124.76 J
Q_0 = mC_f = 39*334 = 13026 J
Q_0-100 = mC_wΔT = 39*4.18*100 = 16302 J
Q_100 = mC_v = 39*2230 = 86970 J
Q_100-108 = mC_sΔT = 39*2.03*(108-100) = 39*2.03*8 = 633.36J

Q = Summation of all the heats added = 1124.76+13026+16302+86970+633.36 = 118056.12 J ≈ 118.06 kJ

7 0

3 years ago

If the same strength force is exerted on two objects why might they be affected differently?​

If the same strength force is exerted on two objects why might they be affected differently?