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Radda [10]
3 years ago
15

Help not to sure with this one need help plz asap​

Physics
2 answers:
katen-ka-za [31]3 years ago
8 0
25,200 meters is the answer
DiKsa [7]3 years ago
4 0

A is the answer for the problem

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Which excerpt are you talking about?
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A baseball is hit with a bat. The direction of the ball is completely reversed and its speed is doubled. If the actual contact w
alexdok [17]
Before the impact, let the velocity of the baseball was v m/s.

After being hit by the bat its velocity is -2v
So, change in velocity, Deltav=v-(-2v)=3v
Acceleration is defined as the rate of change in velocity, i.e. actual change in velocity divided by the time taken to change it. Time taken to change velocity is the time of actual contact of the bat and ball, i.e. 0.31 s.

a=(Deltav)/(Deltat)
=(3v)/0.37
Therefore, a/v=3/0.31=9.7 s^-1
So, the ratio of acceleration of the baseball to its original velocity is 9.7.
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Is rose quartz igneous sedimentary or metamorphic
Dmitriy789 [7]
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7 0
3 years ago
Considering this analogy (current = rate that heat flows through a window, and voltage = temperature difference across the windo
erastovalidia [21]

Answer:

Explained

Explanation:

Resistance R in a current flow through an object is given by

R= \frac{\rho L}{A}

ρ = resistivity of the material

L= length of the object

A=  area of cross section

clearly resistance is directly dependent on length of the object.This means greater the length larger will be resistance to current.

thermal resistance R_th is given by

R_{th} = \frac{L}{KA}

L= length of the object

A=  area of cross section

K = Conductivity of the material

thermal resistance is also is directly dependent on length of the object.This means greater the length larger will be resistance to current.

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On a Saturday afternoon, you decide to pay a neighborhood kid to mow your lawn. The kid usesa manual push lawn mower with a mass
Mars2501 [29]

Answer with Explanation:

We are given that

A.Mass,m=12 kg

\theta=53^{\circ}

\mu_k=0.16

Speed,v=1.5m/s

Net force in x direction must be zero

F_{net}=0

Fsin\theta-f=0

Fsin\theta=f

Net force in y direction

N-mg-Fcos\theta=0

N=mg+Fcos\theta

f=\mu_kN=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_k(mg+Fcos\theta)

Fsin\theta=\mu_kmg+\mu_kFcos\theta

Fsin\theta-\mu_kFcos\theta=\mu_kmg

F(sin\theta-\mu_kcos\theta)=\mu_kmg

F=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}

Power,P=Fv

P=\frac{\mu_kmg}{sin\theta-\mu_kcos\theta}v

Where g=9.8m/s^2

B.Substitute the values

P=\frac{0.16\times 12\times 9.8}{sin53-0.16cos53}\times 1.5

P=40.17W

6 0
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