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Mrac [35]
3 years ago
14

120mL=____cm3-I really need help!

Physics
1 answer:
Ivahew [28]3 years ago
4 0

Answer:

120 mL = 120 cm^3

Explanation:

1 mL is <em>equal</em> to exactly 1 cubic centimeter, so 120 mL = 120 cm^3

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What can lead to groundwater shortages
Alex

Answer:

most commonly occurs because of the frequent pumping of water from the ground.

Explanation:

8 0
3 years ago
During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc
Lina20 [59]
V o = 6 m/s,
t = 2 s
v = 10 m/s
v = v o + a t
a t = v - v o
a = ( v - v o ) / t 
a = ( 10 m/s - 6 m/s ) / 2 s = 4 m/s / 2 s = 2 m/s²
Answer:
The runner`s acceleration is 2 m/s².
6 0
2 years ago
A student uses a spring loaded launcher to launch a marble vertically in the air. The mass of the marble is
GarryVolchara [31]

Answer:

Part a)

When spring compressed by 2 cm

H = 1.47 m

Part b)

When spring is compressed by 4 cm

H = 5.94 m

Explanation:

Part a)

As we know that the spring is compressed and released

so here spring potential energy is converted into gravitational potential energy at its maximum height

So we will have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.02)^2 = 0.003(9.81)(H + 0.02)

so we have

H = 1.47 m

Part b)

Similarly when spring is compressed by 4 cm

then we have

\frac{1}{2}kx^2 = mg(H + x)

0.5(220)(0.04)^2 = 0.003(9.81)(H + 0.04)

so we have

H = 5.94 m

8 0
3 years ago
A client has developed dystrophic calcification as a result of macroscopic deposition of calcium salts. The tissue that would be
Komok [63]

Answer:

Tissues that are damaged or injured.

Explanation:

Dystrophic calcification involves the deposition of calcium in soft tissues despite no disturbance in the calcium metabolism, and this is often seen at damaged tissues.

Examples of areas in the body where dystrophic calcification can occur include atherosclerotic plaques and damaged heart valves.

4 0
3 years ago
A small 0.13 kg metal ball is tied to a very light (essentially massless) string that is 0.70 m long. The string is attached to
beks73 [17]

Answer:

Explanation:

Let l be th length of pendulum

loss of height

= mg ( l - l cos50)

= mg l ( 1-cos50)

1/2 mv² = mgl ( 1-cos50)

v = √[2gl( 1- cos50)]

= √( 2 x 9.8 x .7 x ( 1-cos50)

= 2.2 m / s

speed at the bottom = 2.2 m /s

b )

centripetal acceleration

= v² / r

= 2.2 x 2.2 / .7

= 6.9 m /s²

C )

If T be the tension

T - mg = mv² / r

T = mg + mv² / r

= .13 X 9.8 + .13 X 6.9

= 2.17 N

7 0
3 years ago
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