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OverLord2011 [107]

2 years ago

6

If a car is accelerating downhill under a net force of 3674 N, what force must the brakes exert to cause the car to have constan

t velocity? a no force applied b a greater force c an equal force d a lesser force

1 answer:

nlexa [21]2 years ago

5 0

Answer:

Explanation:

hi

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if you stretch back a rubber band and release it, it shoots across the room. What type of energy conversion has occurr?

matrenka [14]

I believe the answer is potential energy if i remember correctly.

8 0

3 years ago

Read 2 more answers

As an object falls freely toward the earth, the momentum of the object-earth system (1) decreases (2) increases (3) remains the

Brums [2.3K]

The best and most correct answer among the choices provided by your question is the third choice or number 3.

<span>As an object falls freely toward the earth, the momentum of the object-earth system remains the same.</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

4 0

3 years ago

The cartesian coordinates of a point in the xy plane are x = −6.46 m, y = −3.78 m. Find the distance r from the point to the ori

11111nata11111 [884]

Explanation:

Use Pythagorean theorem:

r² = x² + y²

r² = (-6.46 m)² + (-3.78 m)²

r = 7.48 m

5 0

3 years ago

A ball is thrown vertically upwards with a velocity of 9.8 m/s. Its velocity after 1 second will be

Reil [10]

$\LARGE{ \underline{\underline{ \purple{ \bf{Required \: answer:}}}}}$

GiveN:

Initial velocity = 9.8 m/s²
Accleration due to gravity = -9.8 m/s²
Time taken = 1 s

To FinD:

Final velocity of the ball?

Step-by-step Explanation:

Using the first Equation of motion,

⇒ v = u + gt

⇒ v = 9.8 + -9.8(1)

⇒ v = 0 m/s

The final velocity is hence <u>0</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>

<h3>Note:</h3>

While solving questions of under gravity motions using equations of motion, remember the sign convection to avoid mistakes.
You can consider positive above the ground and negative for towards it.

5 0

2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago