How to find angular velocity of an object traveling at a constant speed.

A mobile phone is pulled northward by a force of 10 n and at the same time pulled southward by another force of 15 n. the result

makvit [3.9K]

5 Southward would be the correct answer.

6 0

3 years ago

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

$Workdone = force * distance$

Making "distance" the subject of formula, we have;

$Distance = \frac {workdone}{force}$

Substituting into the equation, we have;

$Distance = \frac {195}{56.7}$

Distance = 3.4 meters.

8 0

3 years ago

G to ml,density=3.291 g/ml

olya-2409 [2.1K]

The density value 3.291 g/ml (gram / milliliter) in words is "three point two nine one g/ml (gram / milliliter)".
=)

7 0

3 years ago

Which liquid will evaporate the fastest water salt water pepsi and gatorade?

MArishka [77]

The liquid that will evaporate the quickest would be water.
<em>~The rest are mixtures, and have more ingrediants in them, therefore the answer should be water. (in which it has no other ingrediants in it but water itself)</em>

6 0

3 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago