All categories

4 years ago

Which of the following describes velocity only in terms of metric units?

1 answer:

8 0

Choice 'D' describes speed in the metric units 'meter' and 'second'.
With no mention of direction, it can't be called 'velocity'.

You might be interested in

over a human lifetime, why don't we easily notice a change in the position of a particular star? A its far away B it moves too s

melamori03 [73]

Answer: A~ It's very far away

Explanation:

5 0

3 years ago

A water balloon is thrown horizontally at a speed of 2.00 m/s from the roof of a building that is 6.00 m above the ground. At th

Elis [28]

Answer:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first, 0.19 s before the other ballon.

Explanation:

The motions of the two water ballons are ruled by the kinematic equations:

$y=y_0+V_0t-gt^2/2$

We are only interested in the vertical motion, so that equation is all what you need.

1. Water ballon is thrown horizontally at sped 2.00 m/s.

The time the ballon takes to hit the ground is independent of the horizontal speed.

Since 2.00 m/s is a horizontal speed, you take the initial vertical speed equal to 0.

Then:

$y=y_0+V_0t-gt^2/2\\ \\ 0=6.00m-9.8\frac{m}{s^2} t^2/2\\ \\ t=\sqrt{2\times6.00m/9.8\frac{m}{s^2}}\\\\ t=1.11s$

2. Water ballon thrown straight down at 2.00 m/s

Now the initial vertical speed is 2.00 m/s down. So, the equation is:

$0=6.00m-2.00\frac{m}{s}t-9.8\frac{m}{s^2}t^2/2\\ \\ 4.9t^2+2t-6=0\\ \\ t=0.92s$

To solve the equation you can use the quadratic formula.

$t=\frac{-2+/-\sqrt{2^2-4(4.9)(-6)} }{2(4.9)}\\ \\ t=-1.33\\ \\ t=0.92$

You get two times. One of the times is negative, thus it does not have physical meaning.

3. Conclusion:

The water ballon that was thrown straight down at 2.00 m/s hits the ground first by 1.11 s - 0.92s = 0.19 s.

5 0

3 years ago

One alternative to burning non-renewable fuels, which harm the environment, could be to burn?

Rashid [163]

Answer:

The answer to that is C. natural gas

3 0

3 years ago

Read 2 more answers

An object is moving at 4.66 m/s when it accelerates at 5.66 m/s2 for a period of 2.35 s. The distance it moves in this time is _

valentinak56 [21]

Answer:

S = 26.58 meters

Explanation:

Given the following data;

Initial velocity = 4.66 m/s

Acceleration = 5.66 m/s²

Time = 2.35 seconds

To find the distance travelled by the object, we would use the second equation of motion;

S = ut + ½at²

Where;

S represents the displacement or height measured in meters.

u represents the initial velocity measured in meters per seconds.

t represents the time measured in seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

S = 4.66*2.35 + ½*5.66*2.35²

S = 10.951 + (2.83 * 5.5225)

S = 10.951 + 15.629

S = 26.58 meters

6 0

3 years ago

Suppose A=BnCm, where A has dimensions LT, B has dimensions L2T-1, and C has dimensions LT2. Then the exponents n and m have the

julsineya [31]

Explanation:

The expression is :

$A=B^nC^m$

A =[LT], B=[L²T⁻¹], C=[LT²]

Using dimensional of A, B and C in above formula. So,

$A=B^nC^m\\\\\ [LT]=[L^2T^{-1}]^n[LT^2}]^m\\\\\ [LT]=L^{2n}T^{-n}L^mT^{2m}\\\\\ [LT]=L^{2n+m}T^{2m-n}$

Comparing the powers both sides,

2n+m=1 ...(1)

2m-n=1 ...(2)

Now, solving equation (1) and (2) we get :

$n=\dfrac{1}{5}\\\\m=\dfrac{3}{5}$

Hence, the correct option is (E).

5 0

3 years ago