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netineya [11]
3 years ago
6

a baseball pitcher throws a fastball at 42 meters per second. if the batter is 18 meters from the pitcher, approximately how muc

h time does it take for the ball to reach the batter?
Physics
2 answers:
lapo4ka [179]3 years ago
6 0
T=D/v = 18/42 = 43 seconds
LiRa [457]3 years ago
4 0
T = distance div time = 18/42 = 3/7 seconds = 0.4 seconds

less than half of a second

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A VW beetle goes from zero to 27m/s in 7.6 seconds. What is the acceleration?
Firlakuza [10]
Initial speed(u)=0m/s
Final speed(v)= 27m/s
Time(t)=7.6s
Use the equation of motion: v = u + at
27 = 0 + a(7.6)
27/7.6 = a
a = 3.55 m/s^2  (3 s.f)


7 0
3 years ago
After a 50-kg person steps from a boat onto the shore, the boat moves away with a speed of 0.70 m/s with respect to the shore. I
Svetradugi [14.3K]

Answer:

M=125 kg

v=1.75 m/s

Explanation:

From the law of linear momentum

  P =mv

Case 1     50*V =M* 0.7     equation 1

               50*V =(M+50)* 0.5    equation 2

equating 1 and 2

               M* 0.7=(m+50)* 0.5

               0.2 M= 25

                    M=125 kg

Putting value of M in equation 1

               50*V =125*0.7

                     V=1.75 m/s

                   

7 0
3 years ago
Read 2 more answers
What is crossing-over in geneitcs and why is it important
slava [35]
Crossing over, or recombination, is the exchange of chromosome segments between nonsister chromatids in meiosis. Crossing over creates new combinations of genes in the gametes that are not found in either parent, contributing to genetic diversity.
8 0
3 years ago
A net force of 25N causes an object to accelerate at 4m/s^2. what is the mass of the object?
Doss [256]

Answer:

6.25 kg

Explanation:

Fnet=ma

m=Fnet/a

m=25/4

m=6.25kg

3 0
3 years ago
What is the velocity of a proton that has been accelerated by a potential difference of 15 kV? (i)\:\:\:\:\:9.5\:\times\:10^5\:m
andre [41]

Answer:

Velocity of a proton, v=1.7\times 10^6\ m/s    

Explanation:

It is given that,

Potential difference, V=15\ kV=15\times 10^3\ V

Let v is the velocity of a proton that has been accelerated by a potential difference of 15 kV.

Using the conservation of energy as :

\dfrac{1}{2}mv^2=qV

q is the charge of proton

m is the mass of proton

v=\sqrt{\dfrac{2qV}{m}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\ C\times 15\times 10^3\ V}{1.67\times 10^{-27}\ kg}

v=1695361.75\ m/s

v=1.69\times 10^6\ m/s

or

v=1.7\times 10^6\ m/s

So, the velocity of a proton is 1.7\times 10^6\ m/s. Hence, this is the required solution.

8 0
4 years ago
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