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Romashka [77]
3 years ago
10

Simple Question! Help!

Physics
2 answers:
Korvikt [17]3 years ago
8 0

<em>60km</em><em>/</em><em>hr</em><em>.</em><em>.</em>

<em>I</em><em> </em><em>think</em><em> </em><em>so</em><em>.</em><em>.</em><em>.</em><em>.</em>

Butoxors [25]3 years ago
3 0
<h3>Answer:</h3>

60 km/s

<h3>step-step wise explaination:</h3>

60 kilometers / 60 minutes

= 1 kilometer / minute

as we know to convert kilometer / min we should multiply 60 to speed value

therefore, (1×60/ 1) km/hr=60 km/ hr

if you want to know why 60 multiplied with it then read this

1 km/min to km/hr

1 minute = 1/60 hour

so, for converting minute to hour we multiply 1/60

when it goes to top then it multiplied

I know it's little complicated let see it by equation

1 km/min = (1 / 1×1/60) km/h

=(1 / 1/60) km/hr

= 60 km/hr

( note there is no change in km so it remains without multiplied)

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(b) A lift (elevator) in a high building transports 12 passengers, each of mass 65 kg, through a
PSYCHO15rus [73]

Explanation:

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I think you can calculate and substitute this units should be watts

5 0
3 years ago
The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to
irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration a_1 for a total time t_1 During this part of the motion, he covers a distance equal to s_1 = 45 m, until he finally reaches a velocity of v_1 = u + a_1t_1. We can use the following suvat equation:

s_1 = u t_1 + \frac{1}{2}a_1t_1^2

which reduces to

s_1 = \frac{1}{2}a_1 t_1^2 (1)

since u = 0.

- In the second part, he continues with constant speed v_1 = a_1 t_1, covering a distance of d_2 = 55 m in a time t_2. This part of the motion is a uniform motion, so we can use the equation

s_2 = v_1 t_2 = a_1 t_1 t_2 (2)

We also know that the total time is 10.0 s, so

t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)

Therefore substituting into the 2nd equation

s_2 = a_1 t_1 (10-t_1)

From eq.(1) we find

a_1 = \frac{2s_1}{t_1^2} (3)

And substituting into (2)

s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1

Solving for t,

s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s

So from (3) we find the acceleration in the first phase:

a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2

And so the average force exerted on the sprinter is

F=ma=(66 kg)(2.34 m/s^2)=154.5 N

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

v_1 = u +a_1 t_1

where we have

u = 0

a_1  =2.34 m/s^2 is the acceleration

t_1 = 6.2 s is the time of the first part

Solving the equation,

v_1 = 0 +(2.34)(6.2)=14.5 m/s

3 0
3 years ago
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