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3 years ago

Explain why the brakes of a car get much hotter than the brakes of a bicycle?

1 answer:

8 0

Answer: Car brakes produces more energy then the bicycle because the cars wheels produces a much bigger force that makes the car go and to stop that force the car uses greater amount of energy that transfers to heat but in a bicycle the wheels do not turn that fast so when you press the break there is less energy that transfer to heat.

Explanation:

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d. The force is doubled and the object’s mass is halved? 18. ||| A man pulling an empty wagon causes it to accelerate at 1.4 m/s

Harrizon [31]

Answer:

$a' = 0.35 m/s^2$

Explanation:

Let say the empty wagon has mass "M"

now by newton's II Law we will have

$F = Ma$

now it is given that empty wagon is pulled with acceleration 1.4 m/s/s

now we will have

$F = 1.4 M$

now a child of mass three times the mass of wagon is sitting on the empty wagon

so here we have

$F = (M + 3M) a$

$1.4 M = 4M a'$

so we have

$a' = 0.35 m/s^2$

8 0

3 years ago

A uniform horizontal rod of mass Mand length rotates with angular velocity about a vertical axis through its center. Attached to

Lostsunrise [7]

Answer:

Explanation:

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3 years ago

An 100W light bulb is on 6 hours work out the energy (kWh) used and the cost of it (1kWh = £0.1359) with working out please

Ainat [17]

1,000 W = 1 kW
100 W = 0.1 kW

(0.1 kW) x (6 h) = 0.6 kWh <=== energy

(0.6 kWh) x (£0.1359/kWh) = £0.0815 <=== cost of it

8 0

3 years ago

The Sun is an enormous ball of gas. Left to itself, a ball of so many atoms should collapse under its own tremendous gravity. Wh

ikadub [295]

Answer:

Nuclear fusion in the Sun's core causes the release of tremendous amounts of energy that leads to very high temperatures and pressure which is much hotter and higher than the temperature and pressure at the exterior surface of the Sun causing the particles in the inner core region to push outwards towards the Sun's surface

Explanation:

3 0

3 years ago

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma

mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

brainly.com/question/15188750

6 0

2 years ago