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tekilochka [14]
3 years ago
8

What is the weight of a 5.00 kg object on Earth? Assume g=9.81 m/s^2.

Physics
1 answer:
Softa [21]3 years ago
7 0

<em>weight = (mass) x (gravity)</em>

Weight = (5.00 kg) x (9.81 m/s²)

weight = (5.00 x 9.81) (kg-m/s²)

<em>Weight = 49.05 Newton</em>

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An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e
Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0
3 years ago
Which of the following is a unit of acceleration?
olasank [31]

{\tt{\red{\underline{\underline{\huge{Answer:}}}}}}

\longrightarrow The rate of change of velocity per unit time is called acceleration.

\longrightarrow Its SI unit is m/s².

\huge\boxed{\fcolorbox{blue}{red}{Thank you}}

7 0
2 years ago
a football player kicks a ball with a mass of 0.42kg. The average acceleration of the football was 14.8 m/s2. How much force did
Zarrin [17]

Answer:

6.216 N

Explanation:

As for Newton's second law of motion

F=ma

where F= the acting force

           m=subjected mass

           a= the acceleration

applying F=ma to the football

F=m*a

 =0.42*14.8

 =6.216 N

6.216 N of a force is supplied to the ball

5 0
2 years ago
A reservoir located in the mountain 250 m above sea level flows through a pipe to a hydroelectric plant in a town at sea level.
Pavlova-9 [17]

Answer:

     v₂ = 70 m / s

Explanation:

For this exercise let's use Bernoulli's equation

where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level

 

          P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂

indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute

         ρ g y₁ = ½ ρ v₂²

         v₂ = \sqrt {2g \ y_1}

let's calculate

         v₂ = √( 2 9.8 250)

         v₂ = 70 m / s

6 0
3 years ago
A "moving sidewalk" in an airport terminal moves at 1.0 m/s and is 35.0 m long. If a woman steps on at one end and walks at 1.5
pishuonlain [190]

Answer:

a.14 s

b.70 s

Explanation:

a.Let the sidewalk moving in positive x- direction.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=1.5m/s

The speed of women relative to the ground

v_w=v_s+v=1+1.5=2.5m/s

Distance=35 m

Time=\frac{distance}{speed}

Using the formula

Time taken by women to reach the opposite end if she walks in the same direction the sidewalk is moving=\frac{35}{v_w}=\frac{35}{2.5}=14s

b.If she gets on at the end opposite the end in part (a)

Then, we take displacement negative.

Speed  of sidewalk relative to ground=v_s=1m/s

Speed of women relative to sidewalk=v=-1.5 m/s

The speed of women relative to the ground=v_w=v_s+v=1-1.5=-0.5m/s

Time=\frac{-35}{-0.5}=70 s

Hence, the women takes 70 s to reach the opposite end if she walks in the opposite direction the sidewalk is moving.

3 0
3 years ago
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