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3 years ago

No idea which one is correct..

Physics

2 answers:

larisa86 [58]3 years ago

8 0

It needs the rate of the faster cyclist and the rate of the slower cyclist

Yuri [45]3 years ago

4 0

Answer:

The answer is D.

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>

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A 62 kg skydiver moving at terminal speed falls 50 m in 1 s. What power is the skydiver expending on the air?

IgorLugansk [536]

Answer:

P = 30380 W

Explanation:

given,

mass of the skydiver, m = 62 Kg

distance, s = 50 m

time, t = 1 s

we know,

$Power= \dfrac{work\ done}{time}$

$Power= \dfrac{F.s}{t}$

F = m g

$Power= \dfrac{m g.s}{t}$

inserting all the values

$P = \dfrac{62\times 9.8\times 50}{1}$

P = 30380 W

hence, Power the skydiver expending on the air is 30380 W

3 0

3 years ago

An unknown mass of each of the following substances, initially at 23.0, absorbs 1910 of heat. The final temperature is recorded

babymother [125]

The data not given here is the specific heat capacity of pyrex glass equal to 0.75(J/g °C). In this case, heat is equal to mass x specific heat capacity x temperature rise. Temperature difference is then equal to 1910 J / 0.75 J/g °C / 23 g equal to 110. 75 <span>°C </span>

3 0

3 years ago

Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg

vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

$\omega = \frac{v}{R}$

Where,

$\omega =$ Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

$\alpha = \frac{\omega}{t}$

Where

$\alpha =$ Angular acceleration

$\omega =$ Angular velocity

t = Time

Our values are

$v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})$

$v = 16.67m/s$

$r = 0.25m$

$t=6s$

Replacing at the previous equation we have that the angular velocity is

$\omega = \frac{v}{R}$

$\omega = \frac{ 16.67}{0.25}$

$\omega = 66.67rad/s$

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

$\alpha = \frac{\omega}{t}$

$\alpha = \frac{66.67}{6}$

$\alpha = 11.11rad/s^2$

Therefore the angular acceleration of a point on the outer edge of the tires is $11.11rad/s^2$

5 0

3 years ago

A mass slider m = 0.200 kg rests on a frictionless horizontal air rail connected to a spring with a force constant k = 5.00 N /

Step2247 [10]

Va ser 0.0900 yo creo preo que esta respuesta te ayude

5 0

3 years ago

The equation for the speed of a satellite in a circular orbit around the earth depends on mass. Which mass?

katovenus [111]

<h3><u>Question: </u></h3>

The equation for the speed of a satellite in a circular orbit around the Earth depends on mass. Which mass?

a. The mass of the sun

b. The mass of the satellite

c. The mass of the Earth

<h3><u>Answer:</u></h3>

The equation for the speed of a satellite orbiting in a circular path around the earth depends upon the mass of Earth.

Option c

<h3><u> Explanation: </u></h3>

Any particular body performing circular motion has a centripetal force in picture. In this case of a satellite revolving in a circular orbit around the earth, the necessary centripetal force is provided by the gravitational force between the satellite and earth. Hence $F_{G} = F_{C}$ .