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Slav-nsk [51]
3 years ago
10

. An object’s resistance to change in motion is dependent solely on what quantity?

Physics
2 answers:
Tomtit [17]3 years ago
6 0
It is dependent upon the object's mass. The greater the mass of the object greater will be the inertia of the object, and hence it's resistance to change in motion as well.
yKpoI14uk [10]3 years ago
4 0

Answer : Mass

Explanation : An object's resistance to change in motion is solely dependent on the mass of the quantity. The tendency to resist the change in motion is called as inertia. Which is highly influenced by the factor called "mass" of the object. The mass of the quantity will decide the direction for change in the motion of a particular object.

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An atom is solid material
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d. causes the cell to make more viruses

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3 years ago
How does charging by conduction work?
anzhelika [568]

Answer:

During charging by conduction, both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively. ... In this case, electrons are transferred from the neutral object to the positively charged rod and the sphere becomes charged positively.

6 0
4 years ago
A weather balloon is designed to expand to a maximum radius of 21 m at its working altitude, where the air pressure is 0.030 atm
dezoksy [38]

Answer:

7.65 m

Explanation:

P_1 = Initial pressure = 0.03 atm

P_2 = Final pressure = 1 atm

r_1 = Inital radius = 21 m

V_1 = Intial volume of gas = \frac{4}{3}\pi r_1^3

V_2 = Final volume of gas = \frac{4}{3}\pi r_2^3

T_1 = Initial temperature = 200 K

T_2 = Final temperature = 323 K

From ideal gas law we have

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m

The radius at liftoff is 7.65 m

7 0
3 years ago
The 15 g head of a bobble-head doll oscillates in SHM at a frequency of 4.0 Hz.
Goshia [24]

Answer:

(a) 9.375 N/m

(b) 0.5024 m/s

(c) 0.01 kg/s

Explanation:

mass of head, m = 15 g = 0.015 kg

frequency, f = 4 Hz

Time period, T = 1 / f = 0.25 s

Let k is the spring constant.

(a)

The formula for the time period is

T=2\pi\sqrt{\frac{m}{K}}

0.25=2\times 3.14 \sqrt{\frac{0.015}{K}}

0.04=\sqrt{\frac{0.015}{K}}

K = 9.375 N/m

(b)

Amplitude, A = 2 cm

Let ω is the angular velocity.

Maximum velocity, v = A ω = A x 2πf

v = 0.02 x 2 x 3.14 x 4 = 0.5024 m/s

(c)

Let b is the damping constant.

A(t = 4s) = 0.5 cm

Ao = 2 cm

Using the formula of damping

\frac{A}{A_{0}}=e^{-\frac{bt}{2m}}

\frac{0.5}{2}}=e^{-\frac{b\times 4}{2\times 0.015}}

0.25=e^{-133.3 b}

Taking natural log on both the sides

ln (0.25) = - 133.3 b

- 133.3 b = - 1.386

b = 0.01 kg/s

3 0
3 years ago
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