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3 years ago

A car turns a certain curve of radius 24.98 m with constant linear speed of

Physics

1 answer:

Anastaziya [24]3 years ago

7 0

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

= 3525.19 kg

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A car drives 215km east and then 45km north. What is the magnitude of the cars displacement? Round you answer to nearest whole n

kifflom [539]

Answer: $219.65\ \text{km}$

Explanation:

Given

Car drives 215 km east and then 45 km North

Displacement is East direction is

$\Rightarrow \vec{r_1}=215\hat{i}$

Now, the displacement from that to 45 km North is given by

$\Rightarrow \vec{r_{21}}=45\hat{j}$

Net displacement is $\vec{r_2}$

$\Rightarrow \vec{r_2}=\vec{r_1}+\vec{r_{21}}\\\Rightarrow \vec{r_2}=215\hat{i}+45\hat{j}$

Magnitude of the displacement is

$\Rightarrow \left | r_{2} \right |=\sqrt{\left ( 215 \right )^2+\left ( 45 \right )^2}\\\Rightarrow \left | r_{2} \right |=\sqrt{48250}\\\Rightarrow \left | r_{2} \right |=219.65\ \text{km}$

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3 years ago

The automobile has a weight of 2700 lb and is traveling forward at 4 ft>s when it crashes into the wall. If the impact occurs

alex41 [277]

Answer:

$F_b=153918\, lb.ft.s^{-2}$

Explanation:

Given:

mass, $m=2700 \,lb$

time, $t=0.06\,s$

velocity, $v=4\,ft.s^{-1}$

coefficient of kinetic friction between wheels & pavement, $\mu=0.3$

According to first condition,

$F\times \Delta t=m\times v$

$F\times 0.06= 2700\times 4$

$F=180000\,lb.ft.s^{-2}$

According to second condition,

<u>Magnitude of frictional force (which acts opposite to the direction of motion):</u>

$f=\mu.N$

where N is the normal reaction.

$f=0.3\times 2700\times 32.2$

$f=26082 \,lb.ft.s^{-2}$

Now, the impulsive force on the wall if the brakes were applied during the crash:

$F_b= F-f$

$F_b=180000-26082$

$F_b=153918\, lb.ft.s^{-2}$

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3 years ago

you push a 51 kg box with a force of 485 N. the friction force on the box is 232 N. calculate the acceleration of the crate.

vichka [17]

Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

F = Fₐ - Fₓ

Substituting the given values in the above equation

F = 485 - 232

= 253 N

The acceleration of the crate is given by

a = F/m

= 253 / 51

= 4.96 m/s²

Hence, the acceleration of the crate is, a = 4.96 m/s²

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