Whom ever can spell the sound of a sniff gets 100 poibts personally

The acceleration of the object may be zero and the velocity of the object may not be equal to zero correct or not

Zinaida [17]

Answer:

Acceleration is the change in velocity, not the velocity itself; therefore, an object can have zero velocity but not zero acceleration because the velocity will be changing.

8 0

3 years ago

A 1022kg Caprice car stopped at an intersection is rear-ended by a 1620kg ranger truck moving with a speed of 14.5m/s. If the ca

Alika [10]

Answer:

Explanation:

mass of car, m = 1022 kg

mass of truck, M = 1620 kg

initial velocity of truck, U = 14.5 m/s

initial velocity of car, u = 0 m/s

Let the final velocity of car is v and the final velocity of truck is V.

Collision is elastic, so the coefficient of restitution, e = 1

Use conservation of momentum

initial momentum of car + initial momentum of truck = final momentum of car + final momentum of truck

m x u + M x U = m x v + M x V

0 + 1620 x 14.5 = 1022 v + 1620 V

23490 = 1022 v + 1620 V ..... (1)

Use the formula of coefficient of restitution

$e = \frac{V_{1}-V_{2}}{u_{2}-u_{1}}$

1 (14.5 - 0) = v - V

14.5 = v - V

V = v - 14.5 .... (2)

Put in equation (1)

23490 = 1022 v + 1620 (v - 14.5)

23490 = 1022 v + 1620 v - 23490

46980 = 2642 v

v = 17.8 m/s

Put in equation (2)

V = 17.8 - 14.5

V = 3.3 m/s

Thus, the speed of car is 17.8 m/s and the velocity of truck is 3.3 m/s after collision.

8 0

3 years ago

Please answer ASAP .

Arisa [49]

Answer:

Explanation:

a = gsinθ

a = 9.8sin30

a = 4.9 m/s²

v² = u² + 2as

v² = 0² + 2(4.9)(10)

v² = 98

v = √98 = 9.8994949...

v = 9.9 m/s

8 0

3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

3 years ago

Give an example of hypothesis for an experiment and then identify its dependent and independent variables. Write all the steps o

e-lub [12.9K]

An example of a hypothesis for an experiment might be: “A basketball will bounce higher if there is more air it”

Step one would be to make an observation... “hey, my b-ball doesn’t have much air in it, and it isn’t bouncing ver high”

Step two is to form your hypothesis: “A basketball will bounce higher if there is more air it”

Step three is to test your hypothesis: maybe you want to drop the ball from a certain height, deflate it by some amount and then drop it from that same height again, and record how high the ball bounced each time.

Here the independent variable is how much air is in the basketball (what you want to change) and the dependent variable is how high the b-ball will bounce (what will change as a result of the independent variable)

Step four is to record all of your results and step five is to analyze that data. Does your data support your hypothesis? Why or why not?

You should only test one variable at a time because it is easier to tell why the results are how they are; you only have one cause.

Hope this helps!

6 0

3 years ago