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anzhelika [568]
2 years ago
12

a football player starts from rest and speeds up to a final velocity of 12 m/s. the speeding up takes a total of 6 seconds. what

is the football players acceleration
Physics
1 answer:
gizmo_the_mogwai [7]2 years ago
8 0

Answer:

2 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 12 m/s

Time (t) = 6 s

Acceleration (a) =?

The acceleration of the player can be obtained as follow:

v = u + at

12 = 0 + (a × 6)

12 = 6a

Divide both side by 6

a = 12 / 6

a = 2 m/s²

Thus, the acceleration of the player is 2 m/s²

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Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele
Marta_Voda [28]

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

8 0
2 years ago
Answer the following. (a) What is the surface temperature of Betelgeuse, a red giant star in the constellation of Orion, which r
bagirrra123 [75]

Answer:

(a) T = 2987.6 k

(b) T = 19986.2 k

Explanation:

The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{\lambda_{max}}

where,

T = Radiated surface temperature

\lambda_{max} = peak wavelength

(a)

here,

\lambda_{max} = 970 nm = 9.7 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{9.7\ x\ 10^{-7}\ m}

<u>T = 2987.6 k</u>

(b)

here,

\lambda_{max} = 145 nm = 1.45 x 10⁻⁷ m

Therefore,

T = \frac{0.2898\ x\ 10^{-2}\ m.k}{1.45\ x\ 10^{-7}\ m}

<u>T = 19986.2 k</u>

6 0
2 years ago
What is the answer to this question
leva [86]
The answer is A 0 degrees is the freezing point of water



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7 0
3 years ago
What's the relationship between magnetism and birds?​
neonofarm [45]

Answer:

can detect axis of magnetic field

Explanation:

3 0
2 years ago
Read 2 more answers
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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