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Helga [31]
3 years ago
7

A bicycle rider travels 50.0 km in 2.5 hours. What is the cyclists speed

Physics
1 answer:
krek1111 [17]3 years ago
7 0
Hi,
Recall the formula V=d/t, where V stands for velocity or speed, d stands for distance and t stands for time. By substituting your values you get: V=50.0km/2.5h, which equals to 20km/h.
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M A spherical conductor has a radius of 14.0cm and a charge of 26.0σC . Calculate the electric field and the electric potential
Yuri [45]

The Electric field is E=23400 V/m.

What is electric potential ?

The amount of labor required to convey a unit of electric charge from a reference point to a given place in an electric field is known as the electric potential (also known as the electric field potential, potential drop, or the electrostatic potential). More specifically, it is the energy per unit charge for a test charge that is negligibly disruptive to the field under discussion.

The electric field of a spherical conductor is given by:

{\displaystyle E={\frac {Q}{4\pi \varepsilon _{0}\,r^{2}}}

Here 'E' is the electric field and 'Q' is the electric charge.

therefore putting the values we get,

E=23400 V/m

To learn more about electric field click on the link below:

brainly.com/question/11509296

#SPJ4

4 0
2 years ago
Two carts, one of mass 2m and one of mass m, approach each other with the same speed, v. When the carts collide, they hook toget
Nikitich [7]

Answer:

<em>Second option</em>

Explanation:

<u>Linear Momentum</u>

The linear momentum of an object of mass m and speed v is

P=mv

If two or more objects are interacting in the same axis, the total momentum is

P_t=m_1v_1+m_2v_2+...

Where the speeds must be signed according to a fixed reference

The images show a cart of mass 2m moves to the left with speed v since our reference is positive to the right

P_1=-2mv

The second cart of mass m goes to the right at a speed v

P_2=mv

The total momentum before the impact is

P_t=-2mv+mv=-mv

The total momentum after the collision is negative, both carts will join and go to the left side

The first option shows both carts with the same momentum before the collision and therefore, zero momentum after. It's not correct as we have already proven

The third option shows the 2m cart has a positive greater momentum than the other one. We have proven the 2m car has negative momentum. This option is not correct either

The fourth option shows the two carts keep separated after the collision, which contradicts the condition of the question regarding "they hook together".

The second option is the correct one because the mass m_2 has a negative momentum and then the sum of both masses keeps being negative

3 0
4 years ago
Read 2 more answers
When does a nebula become a star?
pashok25 [27]
When the contracting gas and dust from nebula become so dense and hot that nuclear fusion starts
7 0
3 years ago
Star Twinkle but planet do not why​
V125BC [204]

Answer:The total variation in the amount of light entering our eye is not dectiable therefore planets do not twinkle.

Explanation:Stars twinkle, while planets (usually) shine steadily. Why? Stars twinkle because … they're so far away from Earth that, even through large telescopes, they appear only as pinpoints. ... Planets shine more steadily because … they're closer to Earth and so appear not as pinpoints, but as tiny disks in our sky.As light from a star races through our atmosphere, it bounces and bumps through the different layers, bending the light before you see it. Since the hot and cold layers of air keep moving, the bending of the light changes too, which causes the star's appearance to wobble or twinkle.

7 0
3 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
3 years ago
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