Please help on this one?

A bicyclist travels 4.5 km west, then travels 6.7 km at an angle 27.0 degrees South of West. What is the magnitude of the bicycl

Dimas [21]

Answer:

10.90km

Explanation:

Magnitude of the total displacement is expressed using the equation

d = √dx²+dy²

dx is the horizontal component of the displacement

dy is the vertical component of the displacement

dy = -6.7sin27°

dy = -6.7(0.4539)

dy = -3.042

For the horizontal component of the displacement

dx = -4.5 - 6.7cos27

dx = -4.5 -5.9697

dx = -10.4697

Get the magnitude of the bicyclist's total displacement

Recall that: d = √dx²+dy²

d = √(-3.042)²+(-10.4697)²

d = √9.2538+109.6146

d = √118.8684

d = 10.90km

Hence the magnitude of the bicyclist's total displacement is 10.90km

6 0

2 years ago

The velocity of a car changes from 15 m/s south to 5 m/s south in 2 seconds what is the acceleration of the car

Leviafan [203]

Acceleration = Change in velocity / time taken

= (5 - 15) / 2

= -5 ms^-2

4 0

3 years ago

Read 2 more answers

A ball of mass 0.160 kg is dropped from a height of 2.25 m. When it hits the ground it compresses 0.087 m.

Studentka2010 [4]

A) 6.64 m/s downward

B) 0.026 s

C) -40.9 N

Explanation:

A)

We can solve this problem by using the law of conservation of energy.

In fact, since the total mechanical energy of the ball must be conserved, this means that the initial gravitational potential energy of the ball before the fall is entirely converted into kinetic energy just before it reaches the floor.

So we can write:

$PE=KE\\mgh = \frac{1}{2}mv^2$

where

m = 0.160 kg is the mass of the ball

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

h = 2.25 m is the initial height of the ball

v is the final velocity of the ball before hitting the ground

Solving for v, we find:

$v=\sqrt{2gh}=\sqrt{2(9.8)(2.25)}=6.64 m/s$

And the direction of the velocity is downward.

B)

The motion of the ballduring the collision is a uniformly accelerated motion (= with constant acceleration), so the time of impact can be found by using a suvat equation:

$s=(\frac{u+v}{2})t$

where:

v is the final velocity

u is the initial velocity

s is the displacement of the ball during the impact

t is the time

Here we have:

u = 6.64 m/s is the velocity of the ball before the impact

v = 0 m/s is the final velocity after the impact (assuming it comes to a stop)

s = 0.087 m is the displacement, as the ball compresses by 0.087 m

Therefore, the time of the impact is:

$t=\frac{2s}{u+v}=\frac{2(0.087)}{0+6.64}=0.026 s$

C)

The force exerted by the floor on the ball can be found using the equation:

$F=\frac{\Delta p}{t}$

where

$\Delta p$ is the change in momentum of the ball

t is the time of the impact

The change in momentum can be written as

$\Delta p = m(v-u)$

So the equation can be rewritten as

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

Here we have:

m = 0.160 kg is the mass of the ball

v = 0 is the final velocity

u = 6.64 m/s is the initial velocity

t = 0.026 s is the time of impact

Substituting, we find the force:

$F=\frac{(0.160)(0-6.64)}{0.026}=-40.9 N$

And the sign indicates that the direction of the force is opposite to the direction of motion of the ball.

4 0

3 years ago

Help with these please?

Anuta_ua [19.1K]

E=kq/r^2

q=(E*r^2)/k

q=(.086N/C)(1.7m^2)/(8.99*10^9N*m^2/C^2)

q=2.76*10^-11 C

q=2.8*10^-11 C

5 0

3 years ago

Read 2 more answers

Which of the following would decrease the magnetic field around a wire?

shusha [124]

The correct answer is
A. decreasing the current

In fact, the magnetic field produced by a current carrying wire is given by
 $B(r) = \frac{\mu_0 I}{2 \pi r}$
where
 $\mu_0$ is the vacuum permeability
I is the current in the wire
r is the distance from the wire at which the field is calculated

We see from the formula that the intensity of the field, B, is directly proportional to the current I, so if the current decreases, the magnetic field strength B decreases as well.

4 0

3 years ago