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3 years ago

How long does it take a car travelling at 60 m.p.h to cover 5 miles?

Physics

2 answers:

bixtya [17]3 years ago

8 0

Answer:

2minutes per mile

Explanation:

Not sure if it helps

k0ka [10]3 years ago

6 0

Answer:

Time = 5 minutes

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You serve a volley ball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has_______ energy. Can

Greeley [361]

Kinetic energy because the ball is in motion or moving with energy behind it... kinda like when you shoot a gun, the bullet is fired out of the muzzle with kinetic energy ( Punch ) and the bullet goes through a wall or something. Sorry but my math skills aren't very good to give complex calculations but I would recommend that you maybe talk to some of the top ranking math guys on the website. Maybe they can give you better help...

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4 years ago

A hydraulic lift is designed to lift cars up to 2000 kg in mass.If the lift under the car is 1.0 m by 1.2m and the area of the i

Lerok [7]

Answer:

C) 160 N

Explanation:

$m$ = mass of the car = 2000 kg

Weight of the car is given as

$W = mg \\W = (2000) (9.8)\\W = 19600 N$

$F_{i}$ = Force exerted on the input piston

$A_{i}$ = Area of input piston = 10 cm x 10 cm = 0.1 m x 0.1 m = 0.01 m²

$A_{lift}$ = Area of lift = 1 m x 1.2 m = 1.2 m²

Using Pascal's law

$\frac{F_{i}}{A_{i}} = \frac{W}{A_{lift}}\\\frac{F_{i}}{0.01} = \frac{19600}{1.2}\\F_{i} = 160 N$

6 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s$

$v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s$

Time taken to reach 1180 m is 11.29 seconds

$v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s$

Time the rocket will keep going up after the engines shut off is 13.06 seconds.

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m$

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s$