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3 years ago

Most engines will contain how many cylinders

Physics

2 answers:

tiny-mole [99]3 years ago

8 0

Usually below 12, There isn't a specific number, VERY FEW have more than 12.

Alisiya [41]3 years ago

4 0

I want to say 4 would be the answer from what I looked at and remembered.

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a uniform metre rule of mass 100g balances at the 40cm mark when a mass X is placed at the 10cm mark what is the value of x.pls

Blababa [14]

Answer:

The fulcrum of the metre stick is at the 40 cm mark

100 g * 10 cm = 1000 g-cm clockwise torque

x * 30 cm = 1000 gm-cm = counterclockwise torque for balance

X = 1000 / (40 -10) = 1000 / 30 = 33.3 gm at 10 cm to balance

3 0

3 years ago

calculate the kinetic energy of a car with a mass of 500kg traveling at a velocity of 10m/s^2 north. A) 2,500j B) 25,000j C) 50,

Arte-miy333 [17]

The correct answer would be A) 2,500j

8 0

3 years ago

The ________ of a gas can affect the pressure of a gas.

Diano4ka-milaya [45]

d. none of the above

pressure affects volume not the other way round

5 0

3 years ago

Read 2 more answers

A plane traveling horizontally at 120 m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target

antoniya [11.8K]

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

8 0

3 years ago

Read 2 more answers

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. (a) Determine the time necessary for the

jeka94

(a) 3.35 s

The time needed for the projectile to reach the ground depends only on the vertical motion of the projectile, which is a uniformly accelerated motion with constant acceleration

a = g = -9.8 m/s^2

towards the ground.

The initial height of the projectile is

h = 55.0 m

The vertical position of the projectile at time t is

$y = h + \frac{1}{2}at^2$

By requiring y = 0, we find the time t at which the projectile reaches the position y=0, which corresponds to the ground:

$0 = h + \frac{1}{2}at^2\\t=\sqrt{-\frac{2h}{a}}=\sqrt{-\frac{2(55.0 m)}{(-9.8 m/s^2)}}=3.35 s$

(b) 78.4 m

The distance travelled by the projectile from the base of the building to the point it lands depends only on the horizontal motion.

The horizontal motion is a uniform motion with constant velocity -

The horizontal velocity of the projectile is

$v_x = 23.4 m/s$

the time it takes the projectile to reach the ground is

t = 3.35 s

So, the horizontal distance covered by the projectile is

$d=v_x t = (23.4 m/s)(3.35 s)=78.4 m$

(c) 23.4 m/s, -32.8 m/s

The motion of the projectile consists of two independent motions:

- Along the horizontal direction, it is a uniform motion, so the horizontal velocity is always constant and it is equal to

$v_x = 23.4 m/s$

so this value is also the value of the horizontal velocity just before the projectile reaches the ground.

- Along the vertical direction, the motion is acceleration, so the vertical velocity is given by

$v_y = u_y +at$

where

$u_y = 0$ is the initial vertical velocity

Using

a = g = -9.8 m/s^2

and

t = 3.35 s

We find the vertical velocity of the projectile just before reaching the ground

$v_y = 0 + (-9.8 m/s^2)(3.35 s)=-32.8 m/s$

and the negative sign means it points downward.

3 0

3 years ago