Someone pleaseeeee helppp

5. How much time does it take for a bird flying at a speed of 45 kilometers per hour to travel a

Lyrx [107]

Answer:

$40h$

Explanation:

Use the velocity formula to solve

$v = \frac{d}{t}$

In this question, you are given velocity $v = 45km/h$ , and you are given a distance, $d = 1800km$ . Time in this question is what you'll need to find.

Start by rearranging the velocity formula, to isolate for t.

$v = \frac{d}{t}$

Start by multiplying both sides by t

$v(t) = \frac{d}{t}(t)\\\\vt = d$

Then divide both sides by v.

$vt\frac{1}{v} = d/v\\ \\t = \frac{d}{v}$

Now that you've isolated for time, sub in your values and calculate.

$t = \frac{d}{v} = \frac{1800km}{45km/h} = 40 h$

8 0

2 years ago

1. You get hit with a force of 45 N by a ball with a mass of 0.75 kg. How fast was the

dlinn [17]

Answer:

$60 {ms}^{ - 2}$

Explanation:

As we know,

=》Force = Mass × Acceleration

=》45 N = 0.75 × Acceleration

=》Acceleration = 45 ÷ 0.75

=》Acceleration = 60

hence, the Acceleration of the ball would be. 60 meters per second square

$60m {s}^{ - 2}$

6 0

2 years ago

A basketball player jumps 76cm to get a rebound. How much time does he spend in the top 15cm of the jump (ascent and descent)?

vesna_86 [32]

Answer:

The time for final 15 cm of the jump equals 0.1423 seconds.

Explanation:

The initial velocity required by the basketball player to be able to jump 76 cm can be found using the third equation of kinematics as

$v^2=u^2+2as$

where

'v' is the final velocity of the player

'u' is the initial velocity of the player

'a' is acceleration due to gravity

's' is the height the player jumps

Since the final velocity at the maximum height should be 0 thus applying the values in the above equation we get

$0^2=u^2-2\times 9.81\times 0.76\\\\\therefore u=\sqrt{2\times 9.81\times 0.76}=3.86m/s$

Now the veocity of the palyer after he cover'sthe initial 61 cm of his journey can be similarly found as

$v^{2}=3.86^2-2\times 9.81\times 0.66\\\\\therefore v=\sqrt{3.86^2-2\times 9.81\times 0.66}=1.3966m/s$

Thus the time for the final 15 cm of the jump can be found by the first equation of kinematics as

$v=u+at$

where symbols have the usual meaning

Applying the given values we get

$t=\frac{v-u}{g}\\\\t=\frac{0-1.3966}{-9.81}=0.1423seconds$

4 0

2 years ago

In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom

krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2 (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R = 38.5 x ( v^2 )

77 x 9.8 x 3.65 = 38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) = 71.54

v = 8.4 m/s ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0

3 years ago

Please help . Anybody

Alona [7]

Answer:

1: c

2: a

3: b

sorry if I am wrong

6 0

2 years ago