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adelina 88 [10]
3 years ago
12

onsider the free body diagram. If the sum of the tension forces is equal to the force of gravity, which description BEST applies

? A) A book is at rest on a tabletop. B) A physics student rests a backpack upon one shoulder. C) A girl hangs by both hands, motionless, from a trapeze. D) A girl falls slowly to Earth while strapped to a parachute.
Physics
1 answer:
cluponka [151]3 years ago
5 0
The answer is C) A girl hangs by both hands, motionless, from a trapeze.
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Which of these ideas was part of the earliest model of the atom?
babymother [125]
I think the answer is D
4 0
3 years ago
Read 2 more answers
Two identical charges, 2 m apart, exert forces of magnitude 4 N on each other. The value of each charge is: 1. 9 × 105 C 2. 4.2
lesya692 [45]

Answer:

The value of each charge is 4.22 x 10⁻⁵ C

Explanation:

Given;

distance between the two identical charges, d = 2 m

the force of repulsion between these two charges, F = 4N

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2} \\\\but \ q_1 =q_2,then \ let \ q_1 =q_2 = q\\\\F = \frac{kq^2}{r^2}\\\\q^2 = \frac{Fr^2}{k}\\\\q^2 = \frac{4*2^2}{9*10^9} \\\\q ^2 = 1.7778*10^{-9}\\\\q = \sqrt{1.7778*10^{-9}}\\\\q =4.22 *10^{-5} C\\\\q= q_1=q_2= 4.22 *10^{-5} C

Therefore, the value of each charge is 4.22 x 10⁻⁵ C

7 0
3 years ago
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the
alexandr402 [8]

Hahahahaha. Okay.

So basically , force is equal to mass into acceleration.

F=ma

so when F=ma , we get acceleration=6m/s/s

Force is doubled.

Mass is 1/3 times original.

2F=1/3ma

Now , we rearrange , and we get 6F=ma

So , now for 6 times the original force , we get 6 times the initial acceleration.

So new acceleration = 6*6= 36m/s/s

5 0
3 years ago
A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
Mila [183]

Answer:

11.4 m/s

Explanation:

The expression for the Centripetal acceleration is :

a=\frac{v^2}{R}

Where, a is the accleration

v is the velocity around circumference of circle

R is radius of circle

In the given question,

a = g = Acceleration due to gravity as the car is at top = 9.81\ m/s^2

v = ?

R = 13.2 m

So,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

<u>v = 11.4 m/s</u>

8 0
3 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
3 years ago
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