All categories

3 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time

Physics

1 answer:

Afina-wow [57]3 years ago

5 0

Answer:

t = 12s

Explanation:

Given:

v-initial = 0 m/s

x = 360 m

a = 5.0 m/s^2

Solve:

x = (v-initial)t + 1/2(a*t^2)

360 = 0t + 1/2 (5.0t^2)

360 = 2.5 t^2

144 = t^2

t = sqrt(144) = 12

Therefore, it takes 12 seconds.

You might be interested in

How much energy is required to increase the temperature of

Basile [38]

Explanation:

Q=msdt

= 2 x 128 x (7-4)

=256 x 3

= 768 J

6 0

3 years ago

If Maria lifts a mass of 10kg into a table of 1.25m high. How much gpe has the mass gained

SCORPION-xisa [38]

Answer:

gravitational potential energy of 10 kg mass is 122.625 J

Explanation:

Gravitational Potential energy gain is given as

$U = mgh$

here we know that

m = 10 kg

h = 1.25 m

now from above formula we have

$U = (10)(1.25)(9.81)$

$U = 122.625 J$

So gravitational potential energy of 10 kg mass is 122.625 J

3 0

3 years ago

What is the gravitational potential energy of a rock with the mass of 67 kg if it is sitting on top of a hill .35 kilometers hig

solong [7]

Not to sure but maybe 23.45

8 0

3 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0

4 years ago

Read 2 more answers

7. A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The oncoming water stream has

NNADVOKAT [17]

Answer:

The magnitude of the average force exerted on the water by the blade is 960 N.

Explanation:

Given that,

The mass of water per second that strikes the blade is, $\dfrac{m}{t}=30\ kg/s$

Initial speed of the oncoming stream, u = 16 m/s

Final speed of the outgoing water stream, v = -16 m/s

We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :

$F=\dfrac{\Delta P}{\Delta t}$

$F=\dfrac{m(v-u)}{\Delta t}$

$F=30\ kg/s\times (-16-16)\ m/s$

F = -960 N

So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.

6 0

4 years ago

Please help me it's URGENT (#10 btw and also how do I solve for time​

Please help me it's URGENT (#10 btw and also how do I solve for time