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sladkih [1.3K]
3 years ago
7

A force gives a 5.0 kg object an acceleration of 2.0 m/s 2. The same force would give a 20 kg object an acceleration of _____. 0

.5 m/s 2 2.0 m/s 2 4.9 m/s 2 8.0 m/s 2
Physics
2 answers:
belka [17]3 years ago
8 0

Answer:

The acceleration of the object is 0.5 m/s²

a is correct.

Explanation:

Given that,

Mass of object = 5.0 kg

Acceleration a = 2.0 m/s^2

Using newton's second law

The force is the product of the mass of the object and acceleration.

F = ma

Put the value of m and a

F = 5.0\times2.0

F = 10\ N

Now, The same force would give a 20 kg object

So, Using newton's second law again

F = ma

The acceleration is

a =\dfrac{F}{m}

Put the value of m and F

a = \dfrac{10}{20}

a =0.5 m/s^2

Hence, The acceleration of the object is 0.5 m/s²

Oksi-84 [34.3K]3 years ago
6 0

m = 5 kg

a = 2 m/s²

to find the force that accelerates the 4 kg object @ 2 m/s²

F = ma = 5 kg x 2 m/s² = 10 N

To find what acceleration 10 N would give a 20 kg object

a = F/m = 10 N/20 kg = 0.5 m/s

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580 nm light shines on a double slit with d=0.000125 m. What is the angle of the third dark interference minimum (m=3)?
Dmitry_Shevchenko [17]

Answer:

0.66 degrees

Explanation:

The computation of the angle of the third dark interference is shown below:

The condition of the minima is

Path difference = (2n +1) × \lambda÷ 2

For third minima, n = 2

Now

xd ÷ D = (2 × 2 + 1) × \lambda÷ 2

d tan Q_3 = 5\lambda ÷ 2

tan Q_3 = 5\lambda ÷ 2d

Q_3 = tan^-1 × (5\lambda ÷2d)

= tan^-1 × (5 × 580 × 10^-9) ÷ (2 × 0.000125)

= 0.66 degrees

5 0
3 years ago
An inductor is connected to the terminals of a battery that has an emf of 12.0 V and negligible internal resistance. The current
algol13

Answer:

R = 2481 Ω  

L= 1.67 H

Explanation:

(a) We have an inductor L which has an internal resistance of R. The inductor is connected to a battery with an emf of E = 12.0 V. So this circuit is equivalent to a simple RL circuit. It is given that the current is 4.86 mA at 0.725 ms after the connection is completed and is 6.45 mA after a long time. First we need to find the resistance of the inductor. The current flowing in an RL circuit is given by

i = E/R(1 -e^(-R/L)*t)                                                   (1)

at t --> ∞ the current is the maximum, that is,  

i_max = E/R

solve for R and substitute to get,  

R=  E/i_max

R = 2481 Ω  

(b) To find the inductance we will use i(t = 0.940 ms) = 4.86 mA, solve (1) for L as,  

Rt/L = - In (1  - i/i_max )

Or,  

L = - Rt/In (1  - i/i_max )

substitute with the givens to get,

L = -(2481 Si) (9.40 x 10-4 s)/ In (1  - 4.86/6.45 )  

L= 1.67 H

<u><em>note :</em></u>

<u><em>error maybe in calculation but method is correct</em></u>

8 0
3 years ago
Read 2 more answers
6. The electric field caused by an electron is weakest near the electron.
OLga [1]

6. "The electric field caused by an electron is weakest near the electron" is FALSE.

7. "An electric field becomes weaker as distance from the electron increase" is TRUE.

<u>Explanation:</u>

The "electrical field" covers the electrical charge and exerts, attracts or repels other charges in the field.The electric field caused by an electron is strongest near the electron while it become weak as distance from the electron increase.

The reason behind is, at a point the direction of the field line is at that point the direction of the field. The relative magnitude of the electric field will be proportional to the field line density. The field is strongest where the field lines are near together and when the field lines are at increasing distance the field is weakest.

5 0
3 years ago
The most expensive benefit is usually ______. a. life insurance b. retirement c. health care d. day care
ch4aika [34]
The answer is C. Health Care
3 0
3 years ago
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A particle of charge 3.53×10 ​−8 ​​ C experiences a force of magnitude 6.03×10 ​−6 ​​ N when it is placed in a particular point
Cloud [144]
<h2>Electric field at the location of the charge is 169.97 N/C</h2>

Explanation:

Electric field is the ratio of force and charge.

Force, F = 6 x 10⁻⁶ N

Charge, q = 3.53 x 10⁻⁸ C

We have           

       E=\frac{F}{q}\\\\E=\frac{6\times 10^{-6}}{3.53\times 10^{-8}}\\\\E=169.97N/C

Electric field at the location of the charge is 169.97 N/C

6 0
3 years ago
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