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fomenos
3 years ago
5

How many seconds will it take to travel 3,600 meters if your speed is 90 meters per second?

Physics
1 answer:
klio [65]3 years ago
4 0
40

Because if you divide 3,600 by 90 it is equivalent to 40.

Hope this helps, have a blessed day :)
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A roller coaster, traveling with an initial speed of 21 m/s, decelerates uniformly at -3.5
liberstina [14]

Answer:

S = Vo t + 1/2 a t^2       distance traveled

t = (V2 - V1) / a = (0 - 21) / -3.5 = 6 sec       time to stop

S = 21 * 6 - 3.5 * 6^2 / 2 = 63 m        distance traveled

3 0
1 year ago
A force of 40 N accelerates a 5 block at 6 m/s ^ 2 along a horizontal surface a. What would the block's acceleration be if the s
Bas_tet [7]

Answer:

a = 8 m/s^2, Ffriction = 10 N, μk = 0.205

Explanation:

a. Force = Mass*Acceleration,

(since you didn't add the units..."5 block"....for the mass, I will assume it to be in kg, per SI units)

40 N = 5 kg*acceleration,

a = 40/5 = 8 m/s^2

b. As you know newtons second law (F=m*a) is actually in the form Fnet = m*a. Which means that if the friction force comes into play, it would be Fapplied - Ffriction = m*a.

Fapplied - Ffriction = m*a,

40 - Ffriction = 5*6,

40 - Ffriction = 30,

Ffriction = 40 - 30 = 10 N

c. The coefficient of kinetic friction is calculated by the formula "Ffriction = μk*Fnormal".

10 = μk*Fnormal (Fnormal = m*g = 5*9.8)

10 = μk*49,

μk=10/49 ≈ 0.205

7 0
3 years ago
Dalton proposed the first atomic model. Which of these statements accurately reflected his thinking at that time?. . . . A.. Ato
QveST [7]
A.. Atoms are alike...
4 0
3 years ago
Read 2 more answers
Why are there temperature differences on the moon's surface even though there is no atmosphere present?
nika2105 [10]

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.  

4 0
3 years ago
What is the maximum value of the magnetic field at a<br> distance2.5m from a 100-W light bulb?
MA_775_DIABLO [31]

To solve this problem we will apply the concepts related to the intensity included as the power transferred per unit area, where the area is the perpendicular plane in the direction of energy propagation.

Since the propagation occurs in an area of spherical figure we will have to

I = \frac{P}{A}

I = \frac{P}{4\pi r^2}

Replacing with the given power of the Bulb of 100W and the radius of 2.5m we have that

I = \frac{100}{4\pi (2.5)^2}

I = 1.2738W/m^2

The relation between intensity I and E_{max}

I = \frac{E_max^2}{2\mu_0 c}

Here,

\mu_0 = Permeability constant

c = Speed of light

Rearranging for the Maximum Energy and substituting we have then,

E_{max}^2 = 2I\mu_0 c

E_{max}=\sqrt{2I\mu_0 c }

E_{max} = 2(1.2738)(4\pi*10^{-7})(3*10^8)

E_{max} = 30.982 V/m

Finally the maximum magnetic field is given as the change in the Energy per light speed, that is,

B_{max} = \frac{E_{max}}{c}

B_{max} = \frac{30.982 V /m}{3*10^8}

B_{max} = 1.03275 *10{-7} T

Therefore the maximum value of the magnetic field is B_{max} = 1.03275 *10{-7} T

3 0
3 years ago
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