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ikadub [295]
3 years ago
5

Questlon 2 of 10

Physics
1 answer:
pochemuha3 years ago
3 0

Answer:

the answer is OD which says tge buoyant girce on an object is equal to the weight of th fluid

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What is the formula for acceleration.
Trava [24]
The formula for acceleration, or average acceleration, is a=(vf-vi) divided by (tf-ti) .
a= Acceleration (or average acceleration
vf= Final Velocity
vi= Initial Velocity
tf= Final Time
ti= Initial Time
hope i helped you out!
3 0
3 years ago
What is the speed of a wave with a frequency of 2 Hz and a wavelength of 87 m?
7nadin3 [17]
V = 174 m/s
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8 0
3 years ago
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What is the measure of the force that gravity applies to an object
GarryVolchara [31]

Answer:

Gravity, or gravitation, is a natural phenomenon by which all things with mass or energy-including planets,stars,galaxies, and even light-are brought toward one another.

Explanation:

4 0
3 years ago
A spring with k = 19.5 N/cm is initially stretched 1.39 cm from its equilibrium length. a) How much more energy is needed to fur
lara31 [8.8K]

Answer:

Energy needed = 54.02 J

Explanation:

the Energy in an elastic spring from hookes law is given  as

F= ke , therefore the energy (E) is

E = \frac{1}{2} Ke^{2}

K = 19.5 N/cm

e = 1.39cm

E = \frac{1}{2} x 19.5 x 1.39

E = 13.55 J

The energy to stretch the spring for 6.93cm is

E = \frac{1}{2} x 19.5 x 6.93

E = 67.57 J

The more energy needed for the further stretch is

67.57 - 13.55

Energy needed = 54.02 J

7 0
3 years ago
A 0.050 kg bullet strikes a 5.0 kg stationary wooden block and embeds itself in the block. The block and the bullet fly off toge
natka813 [3]

Answer: 909 m/s

Explanation:

Given

Mass of the bullet, m1 = 0.05 kg

Mass of the wooden block, m2 = 5 kg

Final velocities of the block and bullet, v = 9 m/s

Initial velocity of the bullet v1 = ? m/s

From the question, we would notice that there is just an object (i.e the bullet) moving before the collision. Also, even after the collision between the bullet and wood, the bullet and the wood would move as one object. Thus, we would use the conservation of momentum to solve

m1v1 = (m1 + m2) v, on substituting, we have

0.05 * v1 = (0.05 + 5) * 9

0.05 * v1 = 5.05 * 9

0.05 * v1 = 45.45

v1 = 45.45 / 0.05

v1 = 909 m/s

Thus, the original velocity of the bullet was 909 m/s

3 0
3 years ago
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