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3 years ago

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A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of

the book?
A.
0.45 kilograms
B.
0.90 kilograms
C.
1.35 kilograms
D.
1.5 kilograms
E.
1.8 kilograms

2 answers:

marysya [2.9K]3 years ago

7 0

I believe D is the right answer

malfutka [58]3 years ago

3 0

B. seems like the answer, I'm not sure.

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What objects do balanced forces act on?

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Objects with balanced forces acting on them experience no change in motion, or no acceleration. So these objects could either be stationary at rest or have a constant velocity. These include a hanging object, a floating object, an object on a table that doesn't move, and a car moving at a constant 10 mph

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2 years ago

How to create or build a homemade golf club. is this possible? If so, what suggestion did you find? Do you have an original idea

goldenfox [79]

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2 years ago

A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr

VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

$rR:\\\\E=k\frac{Q}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

$F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}$

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

$Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}$

with this values of a you can use the following formula:

$a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s$

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0

2 years ago

A college student is working on her physics homework in her dorm room. Her room contains a total of 6.0 x 10^26 gas molecules. A

IceJOKER [234]

Answer:

Temperature, T = 3.62 kelvin

Explanation:

It is given that,

Total number of gas molecules, $N=6\times 10^{26}$

Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W

Time taken, t = 6 min = 360 s

Energy of a gas molecules is given by :

$\Delta E =\dfrac{3}{2}NkT$

$T=\dfrac{2E}{3Nk}$ , k is Boltzmann constant

$T=\dfrac{2\times P\times t}{3Nk}$

$T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}$

T = 3.62 K

So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.

4 0

3 years ago