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3 years ago

Question 3

Physics

1 answer:

pickupchik [31]3 years ago

4 0

When a police officer is trying to decide if a driver is speeding, what is his point of reference. The speed limit

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A boat travels north across a river at a velocity of 22 meters/second with respect to the water. The river's velocity is 2.2 met

MArishka [77]

The magnitude of the resultant is

√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .

The direction of the resultant is

tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .

7 0

3 years ago

A Heavy crate applied a force of 1500 N on a 25m2 piston. What force needs to be applied on a 0.8m2 piston to lift the crate?

Aleks04 [339]

25/1500 is equal to 0.8/x
0.8*1500 is equal to 1200
1200/25 is equal to 48 N

7 0

4 years ago

Read 2 more answers

you give a shopping cart a job down the aisle the cart is full of groceries and has a mass of 18 kilograms the cart accelerates

mezya [45]

Force = (mass) x (acceleration)

Force = (18 kg) x (3 m/s²) = 54 newtons

As long as you continue pushing the cart with 54 newtons of force,
it will accelerate at 3 m/s².

At the instant you release it, or keep your hands on it but stop pushing,
it will stop accelerating. It'll continue forward at the speed it had when
the 54 newtons of force stopped.

6 0

3 years ago

Read 2 more answers

21) A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest

zlopas [31]

Answer:

-22,150 N

Explanation:

When the youngster jumps off the platform, during the fall her initial potential energy is converted into kinetic energy, according to the law of conservation of energy. Therefore, we can write:

$mgh=\frac{1}{2}mu^2$

where the term on the left is the potential energy while the term on the right is the kinetic energy, and where

m = 50.0 kg is the mass of the youngster

$g=9.8 m/s^2$ is the acceleration due to gravity

h = 1.00 m is the heigth of the platform

u is the speed of the youngster as she reaches the floor

Solving for u,

$u=\sqrt{2gh}=\sqrt{2(9.8)(1.00)}=4.43 m/s$

Then, when the youngster hits the floor, the force exerted on her during the deceleration is given by:

$F=\frac{\Delta p}{\Delta t}=\frac{m(v-u)}{\Delta t}$

where $\Delta p$ is her change in momentum, and where

m is the mass

v = 0 is the final velocity (she comes to a stop)

u = 4.43 m/s is the initial velocity

$\Delta t=10.0 ms =0.010 s$ is the duration of the collision

Substituting,

$F=\frac{(50.0)(0-4.43)}{0.010}=-22150 N$

And the negative sign means the direction of the force is opposite to the motion (so, upward).

6 0

4 years ago

An object moving on a horizontal, frictionless surface makes a glancing collision with another object initially at rest on the s

cluponka [151]

Answer:

Momentum is always conserved, and kinetic energy may be conserved.

Explanation:

For an object moving on a horizontal, frictionless surface which makes a glancing collision with another object initially at rest on the surface, the type of collision experienced by this objects can either be elastic or an inelastic collision depending on whether the object sticks together after collision or separates and move with a common velocity after collision.

If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.

Either ways the momentum of the bodies are always conserved since they will always move with a common velocity after collision but their kinetic energy may or may not be conserved after collision, it all depends whether they separates or stick together after collision and since we are not told in question whether or not they separate, we can conclude that their kinetic energy "may" be conserved.

6 0

3 years ago