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densk [106]
3 years ago
15

Matter's resistance to a change in motion is called _____ and is directly proportional to the mass of an object. For an object t

o change its state of motion, a force must be applied to it.
A.Velocity
B.inertia
C.distance
D.area

Physics
1 answer:
sergeinik [125]3 years ago
4 0
The correct answer is B (Inertia)

We define inertia as the resistance to change of the state of motion. That is, a body tends to remain in its state of motion or rests unless an external for acts on it. The quantity depends on the mass of the body. If the body is heavy, it requires a large force to overcome the inertia. 
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if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be
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Answer:

I_{rms}=2.12\ A

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

I=3\sin\theta\ A

We need to find the rms value of the current.

The general equation of current is given by :

I=I_o\sin\theta

It means, I_o=3\ A

We know that,

I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A

So, the rms value of current is 2.12 A.

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A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1
Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain d=2.46\times10^{3}\ m

Height of islandh=1.80\times10^{3}\ m

Distance of the enemy ship from the mountain d'=6.10\times10^{2}\ m

Initial velocity v=2.55\times10^{2}\ m/s

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

v_{x}=v\cos\theta

Put the value into the formula

v_{x}=2.55\times10^{2}\cos74.9

v_{x}=66.42\ m/s

We need to calculate the vertical component of initial velocity

Using formula of vertical component

v_{y}=v\sin\theta

Put the value into the formula

v_{y}=2.55\times10^{2}\sin74.9

v_{y}=246.19\ m/s

We need to calculate the time

Using formula of time

t=\dfrac{d}{v_{x}}

t=\dfrac{2.46\times10^{3}}{66.42}

t=37.03\ sec

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

H= v_{y}t-\dfrac{1}{2}gt^2

Put the value in the equation

H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2

H=2397.4\ m

We need to calculate the distance close to the peak

Using formula of distance

H'=H-h

Put the value into the formula

H'=2397.4-1800

H'=597.4\ m

Hence, The distance close to the peak is 597.4 m.

6 0
3 years ago
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