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Tanzania [10]
3 years ago
7

Two objects are interacting but stay stationary. Which best describes what is happening to he action and reaction forces

Physics
2 answers:
weeeeeb [17]3 years ago
4 0

Answer: B

Explanation: The forces are equal and opposite each other.

Elena L [17]3 years ago
3 0

Answer:

The forces are equal and opposite each other

Explanation:

If two objects are interacting then a force is being applied to them, but because the objects are staying stationary the objects must be exerting the same amount of force but in opposite directions.

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3 years ago
If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a
IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2} (1)

x=V_{o} cos \theta t (2)

Where:

y=0 m is the ball final height (when it hits the ground)

y_{o}=1.1 m is the ball initial height

V_{o}=2.2 m/s is the initial velocity

\theta=30\° is the angle at which the ball was launched

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the horizontal distance the ball travels

Rewriting (1) with the given values:

0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2} (3)

Multiplying all the eqquation by -1 and rearranging:

4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0 (4)

So, since we have a quadratic equation here (in the form of0=at^{2}+bt+c,  we will use the quadratic formula to find  t:  

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}   (5)

Where a=4.9, b=-1.1, c=-1.1  

Substituting the known values and choosing the positive result of the equation, we have:  

t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}  

t=0.59 s (6)

Now, substituting (6) in (2):

x=(2.2 m/s)(cos 30\°)(0.59 s) (7)

x=1.12 m (8) This is the horizontal distance at which the ball hits the ground.

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4 years ago
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Snezhnost [94]
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3 years ago
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If an object sinks in a fluid, the buoyant force is ____ the weight of the object.
olga nikolaevna [1]
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4 years ago
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A bat emits a sound at a frequency of 30.0 kHz as it approaches a wall. The bat detects a beat frequency of 700 Hz. The speed of
ASHA 777 [7]

Answer:

3.948m/s

Explanation:

To solve this we need to apply Doppler effect theory

So

To find the frequency received by insect will be gotten when the Source and observer both are moving in same direction which is given by

f1 = f0 x (V - Vo)/(V - Vs)

f0 = 30.0 kHz

V = 344 m/s

Vs will now be the speed of the bat and

Vo will be the speed of the object which is = 0 m/s

So substituting we have

f1 = 30 x 10^3 x (344- 0)/(344- Vs)

Next to find the frequency reflected by wall we use

f2 = f1 x (V + Vs)/(V + Vo)

So substituting the value of f1 calculated above we have

f2 = 30 x 10^3 x (344 + Vs) x (344 - 0)/[(344 - Vs) x (344 + 0)]

f2 = 30 x 10^3 x (344 + Vs)/(344- Vs)

But the beat frequency detected by bat is 700 Hz,

So we say

f2 - f0 = 700 Hz

30 x 10^3 x (344+ Vs)/(344 - Vs) - 30x 10^3 = 700

(344 + Vs)/(344 - Vs) = 1 + 700/30000 = 1.023

344 + Vs = 344 x 1.023 - Vs x 1.0233

Vs = 344 x ( 1.023 - 1)/(1 + 1.023)

So finally

Vs = Speed of source that is the bat is = 3.949m/s

6 0
4 years ago
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