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3 years ago

You biked to the store in 10 minutes. The store was 3 km away. What was your average speed?

Physics

1 answer:

sineoko [7]3 years ago

4 0

Answer:

speed in km/min:

0.3km/min

speed in km/h:

18km/h

speed in m/s:

5m/s

Explanation:

the average speed is defined as:

$s=\frac{d}{t}$

where $s$ is the speed, $d$ is the distance traveled, and $t$ is the time

according to the statement:

$d=3km = 3,000m$

and

$t=10min=\frac{1}{6}h=600s$ (10 minutes are equal to a sixth of an hour which is 600 seconds)

Because the units in which the speed is required are not indicated, here are some options for the answer:

Depending on the units in which you need the speed, are the quantities you should use.

Speed in km / h (kilometers per hour):

$s=\frac{3km}{\frac{1}{6}h }\\ \\s=18km/h$

speed in m / s (meters per second):

$s=\frac{3,000m}{600s}\\ s=5m/s$

or the speed in km/min (kilometers per minute)

$s=\frac{3km}{10min}\\ s=0.3km/min$

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PLEASE NEED HELP What is the net force acting on the race car in the picture: Question 1 options: 10 N to the right 3 N to the l

Nezavi [6.7K]

Use Net Force Equation.

Fnet = F1 + F2 + F3 + F4

In which the following main forces in play are:

F1 = Gravitational Force

F2 = Normal Force

F3 = Thrust Force

F4 = Air resistance

Since this is assumed to happen on a flat surface, the gravitational and normal forces cancel out. So we are left with:

Fnet = F(thrust) + F(air resistance)

Now we can use this equation to solve for the net force.

Fnet = 10N [right] + 7N [left]

Fnet = 10N [right] - 7N [right]

Fnet = 3N [right].

Therefore, the total net force acting on this car would be 3N [right]

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3 years ago

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

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