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3 years ago

Please help! I’ll give Brainliest

Physics

2 answers:

babymother [125]3 years ago

5 0

Answer:

9.8 secs

Explanation:

the ball is in the air so it takes 9.8 secs to get to the ground

UNO [17]3 years ago

5 0

9.8 is correct because that’s how long it takes the ball to hit the ground

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A convex lens can produce a real image but not a viral image<br> a. true<br> b. false

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The answer is true a convex lens can produce a real image but not a viral image

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4 years ago

4.) mi/hr/s and m/s/s are units for

egoroff_w [7]

Answer:

Acceleration

Explanation:

Acceleration has units of length per time squared.

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3 years ago

The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic

Gekata [30.6K]

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is = $\frac{0+40}{2\\}$ {initial velocity + final velocity}

= $\frac{40}{2}$

=20

Therefore, the magnitude of the average velocity of the car is 20 m/s

3 0

3 years ago

About how much of Earth's water is available for humans to use?

lesya692 [45]

Answer:

0.3 %

Explanation:

Earth cleans and replenishes the water supply through the hydrologic cycle. The earth has an abundance of water, but unfortunately, only a small percentage, is even usable by people.

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3 years ago

A softball of mass 0.220 kg that is moving with a speed of 5.5 m/s (in the positive direction) collides head-on and elastically

Elanso [62]

Answer:

The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

Explanation:

Given that,

Mass of softball = 0.220 kg

Speed = 5.5 m/s

(a). We need to calculate the velocity of the target ball

Using conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$0.220\times5.5+m_{2}\times0=0.220\times(-3.9)+m_{2}v_{2}$

$1.21=-0.858+m_{2}v_{2}$

$m_{2}v_{2}=2.068$ ....(I)

The velocity approach is equal to the separation of velocity

$u_{1}-u_{2}=v_{2}-v_{1}$

$5.5-0=v_{2}-(-3.9)$

$v_{2}=1.6\ m/s$

(b). We need to calculate the mass of the target ball

Now, Put the value of v₂ in equation (I)

$m_{2}\times1.6=2.068$

$m_{2}=\dfrac{2.068}{1.6}$

$m_{2}=1.29\ kg$

Hence, The velocity and mass of the target ball are 1.6 m/s and 1.29 kg.

3 0

3 years ago