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3 years ago

13

a.Calculate the average speed (in km/h) of Charlie, who runs to the store 4 kilometers away in 30 minutes. b.Calculate the dista

nce (in km) that Charlie runs if he maintains this average speed for 1 hour

1 answer:

sveta [45]3 years ago

7 0

as per the question charlie runs to the store which is 4 km away

hence the total distance covered [S] is 4 km

he takes 30 minutes to reach the store.

hence the total time taken [t] = 30 minutes=0.5 hour

We have to calculate the average speed.

the average speed[v]= $=\frac{S}{t}$

= $\frac{4 km}{0.5 hour}$

=8 km/hour

then we have to calculate the total distance traveled by charlie in 1 hour.

the distance covered S= $V_{avg} *t$

=8 km/hour ×1 hour

=8 km

Hence the average speed of charlie is 8 km/hour and he covers a distance of 8 km in 1 hour.

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14. a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s. a. how much later does the ball

Burka [1]

The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s

<h3>Motion Under Gravity</h3>

The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.

Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.

a. how much later does the ball hit the ground?

The time can be calculated by considering the vertical component of the motion with the use of formula below.

h = ut + 1/2gt²

Where

Height h = 75 m

Initial velocity u = 0 ( vertical velocity )

Acceleration due to gravity g = 9.8 m/s²

Time t = ?

Substitute all the parameters into the formula

75 = 0 + 1/2 × 9.8 × t²

75 = 4.9t²

t² = 75/4.9

t² = 15.30

t = √15.3

t = 3.9 s

b. how far from the building will it land?

The range can be found by using the formula

R = ut

Where u = 4.6 m/s ( horizontal velocity )

R = 4.6 × 3.9

R = 18 m

c. what is the velocity of the ball just before it hits the ground?

The final velocity will be

v = u + gt

v = 4.6 + 9.8 × 3.9

v = 4.6 + 38.22

v = 42.82 m/s

Therefore, the answers are 3.9 s, 18 m and 42.82 m/s

Learn more about Vertical motion here: brainly.com/question/24230984

#SPJ1

7 0

1 year ago

How does Sonar work?i will mark brainliest pls help

Fittoniya [83]

I don't like the wording of any of the choices on the list.

SONAR generates a short pulse of sound, like a 'peep' or a 'ping',
focused in one direction. If there's a solid object in that direction,
then some of the sound that hits it gets reflected back, toward the
source. The source listens to hear if any of the sound that it sent
out returns to it. If it hears its own 'ping' come back, it measures
the time it took for the sound to go out and come back. That tells
the SONAR equipment that there IS a solid object in that direction,
and also HOW FAR away it is.

RADAR works exactly the same way, except RADAR uses radio waves.

5 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

b)

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

3 years ago

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long)

loris [4]

This question is incomplete, the complete question is;

Two plane mirrors intersect at right angles. A laser beam strikes the first of them at a point 11.5 cm from their point of intersection, as shown in the figure.

For what angle of incidence at the first mirror will this ray strike the midpoint of the second mirror (which is 28.0 cmcm long) after reflecting from the first mirror

Answer: angle of incidence is 39.4°

Explanation:

Given that;

two plain mirrors intersect at right angle (90°)

distance d = 11.5 cm

S = 28.0 cm

Now the angle that the reflection ray males with first the mirror equal theta (∅)

so

tan∅ = (S/2) / d

tan∅ = (28/2) / 11.5

tan∅ = 14 / 11.5

tan∅ = 1.2173

∅ = tan⁻¹ (1.2173)

∅ = 50.6°

so angle of incidence = 90° - ∅

= 90° - 50.6°

= 39.4°

Therefore angle of incidence is 39.4°

5 0

3 years ago

Tom [10]

1) in the opposite direction always
2) drag force is the force that acts in the opposite direction of the applied force (air resistance if a drag force for example)
3) frictions forces between the sole and the ground as the owner walks
4) chalk would glide very easily, so a creaking sound could never be produced, but also one could not write anything on the board!
5) I apologize for my lack of familiarity with the kind of sport but the probable answer is to raise the friction of their hands in order to decrease slipping and get a tighter, more firm grip

6 0

3 years ago