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puteri [66]
2 years ago
5

If an object is moving and the forces acting on it are unbalanced in the direction of motion, the object will...

Physics
2 answers:
Scrat [10]2 years ago
8 0
Answer:
The object will slow down

Explanation: if you are walking towards harsh wind, you will slow down because of the pressure pushing you.
Alex777 [14]2 years ago
6 0
It will slow down because of the unbalance force
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WATS THE BEST ANIME <br>VERY IMPORTANT PLS ANSWER!!!!!!!​
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There isn’t a best anime , but in my opinion any anime on Netflix is good other than DBZ too many episodes and storyline gets messed up.
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3 years ago
Rick shoots a basketball at an angle of 35' from the horizontal. It leaves his hands 7 feet from the ground with a velocity of 2
Korvikt [17]

Given:

The angle of projection of the basketball, θ=35°

The height at which the ball leaves the hand, h=7 ft

The initial velocity of the basketball, v=20 ft/s

To find:

The parametric equations describing the shot.

Explanation:

The range, x of the basketball is given by,

x=v\cos\theta t

On substituting the known values,

\begin{gathered} x=20\times\cos35\degree\times t \\ \implies x=16.4t \end{gathered}

The change in the height, y of the basketball is given by,

y=-v\sin\theta t+\frac{1}{2}gt^2

Where g is the acceleration due to gravity.

On substituting the known values,

\begin{gathered} y=-20\times\sin35\degree\times t+\frac{1}{2}\times32\times t^2 \\ \implies y=-11.5t+16t^2 \end{gathered}

Final answer:

The parametric equations describing the shot are

\begin{gathered} \begin{equation*} x=16.4t \end{equation*} \\ \begin{equation*} y=-11.5t+16t^2 \end{equation*} \end{gathered}

8 0
1 year ago
What two factors affect the rate of acceleration in an object
Leto [7]
Newton's second law states that F=ma so that means force =mass×acceleration force and mass
8 0
3 years ago
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How does the work required to accelerate a particle from 10 m/s to 20 m/s compare to that required to accelerate it from 20 m/s
poizon [28]

To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore

W = \Delta KE

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

Here,

m = mass

v_{f,i} = Velocity (Final and initial)

First case) When the particle goes from 10m/s to 20m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(20)^2 -\frac{1}{2} (m)(10)^2

W_1 = 150(m) J

Second case) When the particle goes from 20m/s to 30m/s

\Delta W = \frac{1}{2} (m)(v_f)^2 -\frac{1}{2} (m)(v_i)^2

\Delta W = \frac{1}{2} (m)(30)^2 -\frac{1}{2} (m)(20)^2

W_1 = 250(m) J

As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.

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3 years ago
Q2. True or False? A force is a push or a pull that acts on
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