The rate at which an object changes position is that object's _____

An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

6 0

3 years ago

An ac generator consists of a coil with 40 turns of wire, each with an area of 0.06 m2 . The coil rotates in a uniform magnetic

mezya [45]

Answer:

d. 332 V

Explanation:

Given;

number of turns in the wire, N = 40 turns

area of the coil, A = 0.06 m²

magnitude of the magnetic field, B = 0.4 T

frequency of the wave, f = 55 Hz

The maximum emf induced in the coil is given by;

E = NBAω

Where;

ω is angular velocity = 2πf

E = NBA(2πf)

E = 40 x 0.4 x 0.06 x (2 x π x 55)

E = 332 V

Therefore, the maximum induced emf in the coil is 332 V.

The correct option is "D"

d. 332 V

7 0

3 years ago

A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.

Viefleur [7K]

Answer:

Explanation:

a )

change in the gravitational potential energy of the bear-Earth system during the slide = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J - F x 11 = 756.9 J

F x 11 = 4851 J - 756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N

4 0

3 years ago

a mass of 0.75 kg is attached to a spring and placed on a horizontal surface. the spring has a spring constant of 180 N/m, and t

Artemon [7]

Answer:

6.57 m/s

Explanation:

First use Hook's Law to determine the F the compressed spring acts on the mass. Hook's Law F=kx; F=force, k=stiffnes of spring (or spring constant), x=displacement

F=kx; F=180(.3) = 54 N

Next from Newton's second law find the acceleration of the mass.

Newton's .2nd law F=ma; a=F/m ; a=54/.75 = 72m/s²

Now use the kinematic equation for velocity (or speed)

v₂²= v₀² + 2a(x₂-x₀); v₂=final velocity; v₀=initial velocity; a=acceleration; x₂=final displacement; x₀=initial displacment.

v₀=0, since the mass is at rest before we release it

a=72 m/s² (from above)

x₀=0 as the start position already compressed

x₂=0.3m (this puts the spring back to it's natural length)

v₂²= 0 + 2(72)(0.3) = 43.2 m²/s²

v₂= $\sqrt{43.2)\\$ = 6.57 m/s

5 0

4 years ago

The figure shows three displacement vectors, which are

aleksley [76]

Answer:

the correct answer is D

Explanation:

In this exercise, the vectors are in the same west-east direction, so we can assume that the positive direction is east and perform the algebraic sum.

R = δ + ε

where

δ = 2.0 m

ε = 7.0 m

the positive sign indicates that it is heading east

R = 2.0 + 7.0

R = 9.0 m

the direction is east

the correct answer is D

8 0

3 years ago

The rate at which an object changes position is that object's ______.