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Deffense [45]
2 years ago
11

A boy throws a steel ball straight up. consider the motion of the ball only after it has left the boy's hand but before it touch

es the ground, and assume that forces exerted by the air are negligible. for these conditions, the force(s acting on the ball is (are
Physics
2 answers:
kap26 [50]2 years ago
3 0
The forces acting on the ball, aside from air friction, would be the force exerted on the ball by the boy when he threw it up, and gravity working against the motion of the ball
sweet-ann [11.9K]2 years ago
3 0
The gravity is working against the movement of the ball
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1)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it r
DENIUS [597]

Answer:

1)

v_{oy}=11.29\ m/s

2)

y=7.39\ m

Explanation:

<u>Projectile Motion</u>

When an object is launched near the Earth's surface forming an angle \theta with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.

The heigh of an object can be computed as

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

Where y_o is the initial height above the ground level, v_{oy} is the vertical component of the initial velocity and t is the time

The y-component of the speed is

v_y=v_{oy}-gt

1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of v_o

The object will reach the maximum height when v_y=0. It allows us to compute the time to reach that point

v_{oy}-gt_m=0

Solving for t_m

\displaystyle t_m=\frac{v_{oy}}{g}

Thus, the maximum heigh is

\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}

We know this value is 8 meters

\displaystyle y_o+\frac{v_{oy}^2}{2g}=8

Solving for v_{oy}

\displaystyle v_{oy}=\sqrt{2g(8-y_o)}

Replacing the known values

\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}

\displaystyle v_{oy}=11.29\ m/s

2) We know at t=1.505 sec the ball is above Julie's head, we can compute

\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}

\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}

\displaystyle y=1.5\ m+16,991\ m-11.098\ m

y=7.39\ m

5 0
2 years ago
A book rests on the shelf of a bookcase. The reaction force to the force of gravity acting on the book is 1. The force of the sh
Hoochie [10]

Answer:

1. The force of the shelf holding the book up.

Explanation:

The free body diagram of the book is as follows:

1 - The weight of the book towards downwards

2 - The normal force that the shelf exerts on the book towards upwards.

Since the book is at rest, these two forces are equal to each other and according to Newton's Third Law the reaction force to the force of gravity is equal but opposite to the weight of the book. This reaction force is the one that holds the book up on the shelf.

6 0
2 years ago
Read 2 more answers
A runner runs 300 m at an average speed of 3.0 m/s. She then runs another 300m at an average
Kaylis [27]

Answer:

B. 4 m/s

Explanation:

v=d/t

Running for 300 m at 3 m/s takes 100 seconds and running at 300 m at 6 m/s takes 50 seconds. 100 s + 50 s = 150 s (total time). Total distance is 600 m, so 600 m/ 150 s = 4 m/s.

3 0
2 years ago
You buy a AA battery in the store, and it is marked 1.5 V. If this marking is strictly accurate, while this battery is fresh its
vitfil [10]

Answer:

Option A is correct.

when it is used in a circuit. its terminal voltage will be less than 1.5 V.

Explanation:

The terminal voltage of the battery when it is in use in circuits drops lower than the 1.5 V rating given to it due to internal resistance.

All batteries give internal resistances when used in circuits. The internal resistance (though very small) is usually modelled as connected in series with the battery. It is due to some form of interference from the chemical makeup of the battery.

Normally, while the battery is fresh, the voltage (V) obtained at its terminals when connected in series with a resistor of resistance R is V = IR; where I is the current flowing in this circuit.

But once the interenal resistance (r) of the battery comes into play,

V = I₁ (r + R)

The current in the circuit evidently drops (that is I₁ < I) and V = (I₁r + I₁R)

The voltage across the terminals of the battery is no longer V but is now (V) × [R/(R+r)] which is less than the initial V and it reduces as the internal resistance, r, increases.

Hope this Helps!!!

3 0
2 years ago
What is the minimum work needed to push a 950-kg car 710 m up along a 9.0° incline? ignore friction?
kakasveta [241]
<span>1.0344645 MJ The minimum energy need is the potential energy of the car at the top of the ramp and is given by mass*gravity*height mass is known, gravity is assumed to be 9.81m/s^2 as it is on earth, and height must be calculated using trigonometry. height=sin(9 degrees)*710m=111meters so potential energy = 950kg*111m*9.81m/s^2=1.0344645 MJ Using the law of the conservation of energy we can assume that the energy expended to push the car up the incline was at least the potential energy gained by moving 111m against the pull of gravity.</span>
7 0
2 years ago
Read 2 more answers
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