Which of the following refers to anything that has mass and takes up space

A girl rides her scooter to school, a total distance of 4.5km. She has to slow down twice to cross busy streets, but overall the

Whitepunk [10]

Answer:

6.93 km/h

Explanation:

To calculate her average speed, we need the "speed" formula, which is:

average speed = distance / time

You plug in your numbers and it will give you the answer.

Speed = 4.5km/0.65hr

= 6.923 km/h

8 0

3 years ago

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"Which of the following best describes the circuit shown below?

Anettt [7]

The figure shown above is series combination as the two resistors (bulb) are there which are then connected to the battery
so i conclude from the above options given the option is B
hope it helps

3 0

3 years ago

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Calculate the east component of a resultant 32.5 m/s, 35.0° east of north.

ValentinkaMS [17]

Answer:

East component is: 18.64 m/s

Explanation:

If the resultant is 32.5 m/s directed 35 degrees east of north, then we use the sin(35) projection to find the east component of the velocity:

East component = 32.5 m/s * sin(35) = 18.64 m/s

4 0

3 years ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

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