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Shkiper50 [21]
3 years ago
10

A motorcycle has a speed of 30 m / s. After braking, it decelerates with constant deceleration A = -3.0 m / s ^ 2

Physics
2 answers:
kenny6666 [7]3 years ago
7 0

Answer:21

Explanation:

marin [14]3 years ago
4 0

a = \displaystyle\frac{v-u}{t}

-3 = \displaystyle\frac{v - 30}{3}

v -30 = -9

v = 21 m/s

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A woman drives 200 miles west and then turns south and drives 375 miles. Her trip takes her 11.75 hrs. What is the woman's avera
FromTheMoon [43]

Answer:

The woman's average velocity during the trip is 36.2 miles/hour.

Explanation:

Velocity can be define as the displacement of an object per time. It is a vector quantity, and measured in m/s.

i.e velocity = \frac{displacement}{time}

From the given question,

Displacement = \sqrt{200^{2} + 375^{2}  }

                       = \sqrt{40000+140625}

                       = \sqrt{180625}

                       = 425

The displacement of woman is 425 miles.

velocity = \frac{425}{11.75}

             = 36.1702 miles/hour

The woman's average velocity during the trip is 36.2 miles/hour.

7 0
3 years ago
Suppose you were asked to find the torque about point p due to the normal force n in terms of given quantities. which method of
igomit [66]

Answer:

T=Lnsin\alpha

Please check the attached

Explanation:

The torque can simply be calculated by multiplying the length of the rod by the perpendicular force n as shown in the attached figure.

Note that sin90=1

T=Lsin\alpha(nsin90)

T=Lsin\alphaxn

T=Lnsin\alpha

7 0
3 years ago
A 1-kg rock is suspended by a massless string from one end of a
maxonik [38]

Answer:

The weight of measuring stick is 9.8 N

Explanation:

given information:

the mass of the rock, m_{r} = 1 kg

measuring stick, x =1 m

d = 0.25 m

to find the weight of measuring stick, we can use the following equation:

τ = Fd

τ = 0

F_{r} d - F_{s}d = 0

F_{r} = the force of the rock

F_{s} = the force of measuring stick

F_{s} =F_{r}

    = m g

    = 1 kg x 9.8 m/s

    = 9.8 N

thus, the weight of measuring stick is 9.8 N

6 0
3 years ago
6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and obser
Andrej [43]

Answer:

Answer is C. Both technicians A and B.

Refer below.

Explanation:

Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:

Both technicians A and B

7 0
3 years ago
Read 2 more answers
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