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motikmotik
2 years ago
14

Imagine a rock is dropped from the top of a tall building. After 2 seconds of falling, the rock’s instantaneous speed is approxi

mately 20m/s. does anyone know is it true or false
Physics
2 answers:
pshichka [43]2 years ago
6 0

Answer:

TRUE

Explanation:

As we know that stone is falling freely from rest

so here we can use kinematics to find the final speed of the stone

v_f = v_i + at

here we know that

v_i = 0

a = 9.81 m/s^2

t = 2 s

now we have

v_f = 0 + (9.81)(2)

v_f = 19.62 m/s

so final speed of the rock is nearly 20 m/s

shepuryov [24]2 years ago
3 0
True
It's in free fall( 9.8 m/s), 2 seconds have passed.
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11. An object moves in circular path with constant speed
Temka [501]

Answer:

B. Is its acceleration constant

Explanation:

Uniform circular motion can be described as the motion of an object in a circle at a constant speed. As an object moves in a circle, it is constantly changing its direction. ... An object undergoing uniform circular motion is moving with a constant speed. Nonetheless, it is accelerating due to its change in direction.

6 0
2 years ago
Which of the following ranching practices contributes to soil erosion?
Natasha2012 [34]

A. overgrazing of livestock

Explanation:

Overgrazing of livestock is the ranch practice that contributes to soil erosion.

Soil erosion is the washing away of the top layer of the earth on which plant grows.

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6 0
3 years ago
A projectile is shot on level ground with a
erastova [34]

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m    agreeing with answer found above

5 0
2 years ago
An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 . The same
Hatshy [7]

Answer:

v_2=\sqrt{2}v_1

Explanation:

The velocity v₁ can be calculated with the kinematic formula:

v_1^{2} =v_0^{2} +2gh

Since the object is initially at rest, v₁ becomes:

v_1=\sqrt{2gh}

Where g is the acceleration due to gravity. Now, the velocity v₂ can be calculated with the same formula, but now the initial velocity is v₁:

v_2^{2}=v_1^{2} +2gh

Substituting v₁ in this expression and solving for v₂, we get:

v_2^{2}=(\sqrt{2gh} )^{2} +2gh=4gh\\\\\implies v_2=\sqrt{4gh}=2\sqrt{gh}

Now, dividing v₂ over v₁, we get the expression:

\frac{v_2}{v_1}=\frac{2\sqrt{gh} }{\sqrt{2gh}}=\sqrt{2}\\   \\\implies v_2=\sqrt{2}v_1

It means that v₂ is √2 times v₁.

4 0
3 years ago
X-rays with frequency 3 ⨉ 10 18 Hz shine on a crystal, producing an interference pattern. If the first bright spot is observed a
FrozenT [24]

Answer:

5.1645\times 10^{-11}\ m

Explanation:

n = Order = 1

c = Speed of light = 3\times 10^8\ m/s

f = Frequency = 3\times 10^{18}\ Hz

\theta = Angle = 75.5^{\circ}

Lattice spacing is given by

d=\dfrac{n\lambda}{2\sin\theta}\\\Rightarrow d=\dfrac{n\times c}{f2\sin\theta}\\\Rightarrow d=\dfrac{1\times 3\times 10^{8}}{3\times 10^{18}\times 2\times \sin75.5^{\circ}}\\\Rightarrow d=5.1645\times 10^{-11}\ m

The lattice spacing of the crystal is 5.1645\times 10^{-11}\ m

6 0
2 years ago
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