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3 years ago

In which of the following situations or places would you want to reduce the

Physics

1 answer:

sashaice [31]3 years ago

4 0

Answer:

Explanation:

The correct answer would be in the axle of the wheels while you ride your bicycle.

Options A, B, and C requires that the forces of friction is increased in order to have more control.

However, option D requires that there is a minimal frictional force in the axle of the wheels of a bicycle while riding so that a little effort would be required to keep the bicycle moving.

The lesser the friction, the lower the effort that would be needed to keep the bicycle moving and vice versa.

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Heat moves from hot to cold either through air or liquid

KonstantinChe [14]

Answer:

both

Explanation:

because when it is hot in summer 5hat is the air and the sund u can warm things up and then it get hot

5 0

3 years ago

The position of a certain airplane during takeoff is given by x=1/2 *bt2, where b = 2.0 m/s2 and t = 0 corresponds to the instan

Serhud [2]

Answer:

1362000 kgm/s

Explanation:

So the total mass combination of the plane and the people inside it is

M = 35000 + 160*65 = 45400 kg

After 15 seconds at an acceleration of 2 m/s2, the plane speed would be

V = 2*15 = 30 m/s

So the magnitude of the plane 15s after brakes are released is

MV = 45400 * 30 = 1362000 kgm/s

5 0

3 years ago

A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?

Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

3 0

3 years ago

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

3 years ago

A car has a mass of 1.00 × 103 kilograms, and it has an acceleration of 4.5 meters/second2. What is the net force on the car?

aksik [14]

ANSWER

C. $F=4.5 \times10^3$ . newtons

EXPLANATION

According to Newton's second law,

$F_{net}=ma$ , where

$m=1.00\times 10^3kg$ is the mass measured in kilograms.

and

$a=4.5ms^{2}$ is the acceleration in metres per second square.

We substitute these values to obtain,

$F=1.00\times10^3 \times 4.5$ .

We rearrange to get,

$F=1.00\times4.5 \times10^3$ .

We multiply out the first two numbers and leave our answer in standard form to get,

$F=4.5 \times10^3 N$ .

The correct answer is C

3 0

4 years ago