A force sets an object in motion when the force is multiplied

A car travels a distance of 320 km in 4 hours. What is your average speed in meters per second?

Andreas93 [3]

Answer:

22.2 m/s

Explanation:

First, we need to convert km to m by multiplying by 1000. This means that the car traveled 320 000 meters.

Next, we convert hours to minutes by multiplying by 3600 (the number of seconds in an hour). This means that overall, the car traveled 320 000 m in 14 400 seconds.

The average speed can be found by using the equation $\frac{distance}{time}$ . After substitution, this gives the fraction $\frac{320 000}{14 400}$ , which reduces to 22 $\frac{2}{9}$ m/s, or about 22.2 m/s.

4 0

3 years ago

A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

A merry-go-round is spinning at a rate of 4.04.0 revolutions per minute. Cora is sitting 0.50.5 m from the center of the merry-g

dsp73

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

$f = \frac{4.04}{60} =0.067 rev/s$

here we know that angular frequency is given as

$\omega = 2\pi f$

$\omega = 2\pi(0.067)$

$\omega = 0.42 rad/s$

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

7 0

3 years ago

Which condition must exist in order for conduction to occur between two substances

motikmotik

Heat!!!!!!!!!!!!!!!!!!! sorry about the exclamation marks but it wont let me post in less it has more than 20 characteristics

6 0

3 years ago

A dart with mass md is launched toward a block of mass mb that is suspended from a string of length L. The dart is moving horizo

Yuki888 [10]

Answer:

A) Impulse is the same for both the objects

B) The higher is the speed, the greater will be the height.

Explanation:

Part a)

The time of interaction of the two bodies i.e the hanging mass and the stick is same. Thus, force caused by dart on the block = force caused by block on the dart. Hence, impulse is the same for both the objects.

Part B

The energy will be conserved in the entire reaction process

Hence, Kinetic energy = potential energy

0.5Mv^2 = gh(md+mb)

H is directly proportional to the square of speed.

Hence, the higher is the speed, the greater will be the height.

5 0

2 years ago