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Grace [21]
3 years ago
11

True or false once a object hits the water the forces are balanced and the object will stop

Physics
2 answers:
SVEN [57.7K]3 years ago
6 0

Answer:

The statement will be false.

Explanation:

Newton's First Law, also called the Law of inertia, indicates that "Every body perseveres in its state of rest or uniform rectilinear motion unless it is forced to change its state by forces printed on it." This means that for a body to leave its state of rest or uniform rectilinear motion, it is necessary for a force to act on it. This is why Newton's first law is known as the Law of Inertia, because inertia is the resistance that an object has to a change in its state of motion.

On the other hand, it is said that the forces are balanced when the net force on the object is zero (that is, the result of the sum of all the forces is zero) and the forces are not balanced when the net force is nonzero (that is, the result of the sum of all forces is not zero).

In Newton's first Law, the forces are balanced while persevering in their state of rest or uniform rectilinear motion. When a force acts on a body, an imbalance occurs as the body leaves its state of rest or uniform rectilinear motion.

Then <em><u>the statement will be false</u></em>. When the forces are balanced the body will remain in its state of rest or uniform rectilinear motion. That is to say that once the object hits the water it will continue moving in the  velocity at which it moved when it hit the water (uniform rectilinear motion) until a force acts on it and forces it to change its state.

sergejj [24]3 years ago
5 0
Answer: False.

Once the object hits the water, the forces are balanced (50 N down and 50 N up). However, an object in motion will continue in motion at the same speed and in the same direction. When the object strikes the water, it stops accelerating; yet it does not stop moving.

I hope this helps! :)
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Racing cars driven by chris and kelly are side by side at the start of a race. the table shows the velocities of each car (in mi
Mamont248 [21]

Solution

distance travelled by Chris

\Delta t=\frac{1}{3600}hr.

X_{c}= [(\frac{21+0}{2})+(\frac{33+21}{2})+(\frac{55+47}{2})+(\frac{63+55}{2})+(\frac{70+63}{2})+(\frac{76+70}{2})+(\frac{82+76}{2})+(\frac{87+82}{2})+(\frac{91+87}{2})]\times\frac{1}{3600}

=\frac{579.5}{3600}=0.161miles

Kelly,

\Delta t=\frac{1}{3600}hr.

X_{k}=[(\frac{24+0}{2})+(\frac{3+24}{2})+(\frac{55+39}{2})+(\frac{62+55}{2})+(\frac{71+62}{2})+(\frac{79+71}{2})+(\frac{85+79}{2})+(\frac{85+92}{2})+(\frac{99+92}{2})+(\frac{103+99}{2})]\times\frac{1}{3600}

=\frac{657.5}{3600}

\Delta X=X_{k}-X_{C}=0.021miles

4 0
3 years ago
How magnitude, distance and velocity affects motion?
levacccp [35]

Answer:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

Explanation:

Velocity is the rate of motion in a specific direction. ... My velocity is 30 kilometers per hour that-a-way. Average speed is described as a measure of distance divided by time. Velocity can be constant, or it can change (acceleration).

6 0
3 years ago
Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an
ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega

Power factor is given by

F=\dfrac{R}{Z}\\\Rightarrow F=\dfrac{280}{972.1483}\\\Rightarrow F=0.28802

The power factor is 0.28802

The average power to the circuit is given by

P=\dfrac{V^2}{Z}\\\Rightarrow P=\dfrac{50^2}{972.1483}\\\Rightarrow P=2.57162\ W

The average power to the circuit is 2.57162 W

Power to resistor

P_R=IR\\\Rightarrow P_R=5.1\times 10^{-2}\times 280\\\Rightarrow P_R=14.28\ W

Power to resistor is 14.28 W

Power to inductor

P_L=IX_L\\\Rightarrow P_L=5.1\times 10^{-2}\times 10500\times 100\times 10^{-3}\\\Rightarrow P_L=53.55\ W

Power to the inductor is 53.55 W

Power to the capacitor

P_C=IX_C\\\Rightarrow P_C=5.1\times 10^{-2}\times \dfrac{1}{10500\times 0.8\times 10^{-6}}\\\Rightarrow P_C=6.07142\ W

The power to the capacitor is 6.07142 W

8 0
3 years ago
When a certain air-filled parallel-plate capacitor is connected across a battery, it acquires a charge of magnitude 172 μC on ea
crimeas [40]

Answer:

k = 2.279

Explanation:

Given:

Magnitude of charge on each plate, Q = 172 μC

Now,

the capacitance, C of a capacitor is given as:

C = Q/V

where,

V is the potential difference

Thus, the capacitance due to the charge of 172 μC will be

C = \frac{(172\ \mu C)}{V}

Now, when the when the additional charge is accumulated

the capacitance (C') will be

C' = \frac{(172+220)\ \mu C)}{V}

or

C' = \frac{(392)\ \mu C)}{V}

now the dielectric constant (k) is given as:

k=\frac{C'}{C}

substituting the values, we get

k=\frac{\frac{(392\ \mu C)}{V}}{\frac{(172)\ \mu C)}{V}}

or

k = 2.279

6 0
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Solar energy, wind energy, and hydroelectric energy are all renewable energy
5 0
4 years ago
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