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3 years ago

8

What is the maximum speed with which a 1200-kg car can round a turn of radius 94.0 m on a flat road if the coefficient of static

friction between tires and road is 0.50?

1 answer:

JulijaS [17]3 years ago

7 0

Answer:

<h2> v= 21.47m/s </h2>

Explanation:

For the car to turn at the about the centripetal force must not be greater than the static friction between the tires and the road

we will use the expression relating centripetal force and static friction below

let U represent the coefficient of static friction

Given that

U= 0.50

mass m= 1200-kg

radius r= 94.0 m

Assuming g= 9.81 m/s^2

$U*m*g=\frac{mv^2}{r}$

$U*g=\frac{v^2}{r}$

substituting our given data in to expression we can solve for the speed V

$0.5*9.81=\frac{v^2}{94}$

making v the subject of formula we have

$0.5*9.81=\frac{v^2}{94}\\\v= \sqrt{0.5*9.81*94} \\\\v= \sqrt{461.07} \\\\v= 21.47$

v= 21.47m/s

<em><u>hence the maximum velocity of the car is 21.47m/s</u></em>

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Two rams run toward each other. One ram has a mass of 49 kg and runs west

olga nikolaevna [1]

Answer: (d)

Explanation:

Given

Mass of the first ram $m_1=49\ kg$

The velocity of this ram is $v_1=-7\ m/s$

Mass of the second ram $m_2=52\ kg$

The velocity of this ram $v_2=9\ m/s$

They combined after the collision

Conserving the momentum

$\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]$

Momentum after the collision will be

$\Rightarrow 101\times \dfrac{125}{101}=125\ kg-m/s\ \text{East}$

Therefore, option (d) is correct

4 0

2 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

Starting velocity: 50 m/s

Taya2010 [7]

Please find attached photograph for your answer. Please do comment whether it is useful or not

6 0

2 years ago

A tennis ball is dropped from 1.13 m above the

Alex73 [517]

Answer:

-4.71 m/s

Explanation:

Given:

y₀ = 1.13 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2a (y − y₀)

v² = (0 m/s)² + 2(-9.8 m/s²) (0 m − 1.13 m)

v = -4.71 m/s

7 0

2 years ago

A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T

sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

$\Delta U = Q-W$

$\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)$

Where

$\dot{m} =$ Mass flow

$h_1 =$ Specific Enthalpy (IN)

$h_2 =$ Specific Enthalpy (OUT)

$g =$ Gravity

$z_{1,2} =$ Heigth state (In, OUT)

$V_{1,2} =$ Velocity (In, Out)

Our values are given by,

$T_i = 10\°C$

$P_1 = 100kPa$

$\dot{m} = 5kg/s$

$z_2 = 20m$

For this problem we know that as pressure, temperature as velocity remains constant, then

$h_1 = h_2$

$V_1 = V_2$

Then we have that our equation now is,

$\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}$

$\dot{W} = \frac{(5)(9.81)(0-20)}{1000}$

$\dot{W} = -0.98kW$

8 0

3 years ago