A 1500 kg car skids to a halt on a wet road where μk = 0.51. how fast was the car traveling if it leaves 66-m-long skid marks?

1 answer:

3 0

We got coefficient of friction uk = 0.51
Mass m = 1500 kg
normal force on the car = weight of the car = 1500 * 9.8 N = 14700 N
and the force of friction on the car, f = uk * normal force = 0.51 * 14700 N = 7497 N
taking Acceleration a = -f/m = -7497/1500 = -4.998 m/s^2
Displacement s = 66 m
Final velocity v = 0
Initial velocity u = ?
v^2 = u^2 + 2as
0 = u^2 - 2 * 4.998 * 66
u = 25.69 m/s

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Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a

Harrizon [31]

Answer:

The pressure exerted by camel feet is 2000 N/m².

Step-by-step explanation:

<h3>Solution :</h3>

Here, we have given that ;

Force applied on camel feet = 4000 N
Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

${\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}$

Substituting all the given values in the formula to find the pressure exerted by camel feet.

$\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\ \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}} \\ \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\ \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\ \\\star \: \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}$