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Diano4ka-milaya [45]
3 years ago
10

2342-24521+2453x=1342

Physics
1 answer:
Dmitry [639]3 years ago
3 0

Answer:

9.59

Explanation:

Given expression:

       2342  - 24521 + 2453x  = 1342

Now,

    add similar terms on both sides of the expression:

              -22179 + 2453x  = 1342

 So;

  Collect like terms;

              2453x  = 1342 + 22179

               2453x  = 23521

 Now, divide both sides by 2453;

              x  = 9.59

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4.6 billion year I'm positive

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A 0.700-kg particle has a speed of 1.90 m/s at point circled A and kinetic energy of 7.20 J at point circled B. (a) What is its
Savatey [412]

Answer:

a). E_{kA}=1.2635 J

b). V_{B}=4.535\frac{m}{s}

c). ΔE_{t}=8.4635 J

Explanation:

ΔE=kinetic energy

a).

E_{kA}=\frac{1}{2}*m*v_{A} ^{2} \\ v_{A}=1.9 \frac{m}{s}\\ m=0.70kg\\E_{kA}=\frac{1}{2}*0.70kg*(1.9 \frac{m}{s})^{2} \\E_{kA}=1.2635 J

b).

E_{kB}=\frac{1}{2}*m*v_{B} ^{2}

V_{B}^{2}=\frac{E_{kB}*2}{m} \\V_{B}=\sqrt{\frac{E_{kB}*2}{m}} \\V_{B}=\sqrt{\frac{7.2J*2}{0.70kg}} \\V_{B}=4.53 \frac{m}{s}

c).

net work= EkA+EkB

E_{t}=E_{kA}+ E_{kB}\\E_{t}=1.2635J+7.2J\\E_{t}=8.4635J

3 0
3 years ago
A race car has a velocity of 382 km/h to the right. If the car’s mass is 705 kg and the driver’s mass is 65 kg, what force is ne
Lady bird [3.3K]

Answer:

Given: Vi = 382 km/h, Vf = 0 km/h, Mc = 705 kg, Md = 65 kg, Δt = 12

Required: Δx

F = Δp / Δt

  = \frac{(Mc+Md)Vf-(Mc+Md)Vi}{t} \\\\= 6.81 * 10x^{3} N [left]\\\\x=\frac{1}{2} (Vi+Vf)\\ \\ = 637m[right]

6 0
3 years ago
(II) How much work did the movers do (horizontally) pushing a 46.0-kg crate 10.3 m across a rough floor without acceleration, if
Gwar [14]

Answer:

2324 J

Explanation:

The formula for work is:

W=F*d

where F is the force applied, and d is the distance moved, in this case d=10.3m

and we need to find F.

Since the crate is not moving up or down, we conclude that the <u>normal force must be equal to the weight </u>of the object:

N=w

where N is the normal force and w is the weight, which is: w=mg, where g is the gravitational acceleration g=9,81m/s^2 and m is the mass m=46kg.

---------

Thus the normal force is:

N=mg

Now, the force due to the friction is defined as:

f=\mu N=\mu mg

where \mu is the coefficient of friction, \mu =0.5

So, for the crate to move, the force applied must be equal to the frictional force:

F=f\\F=\mu mg

And now that we know the force we can calculate the work:

W=F*d\\W=\mu mg*d

substituting known values:

W=(0.5)(46kg)(9.81m/s^2)(10.3)\\W=2324J

5 0
3 years ago
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The mass of the Book is 2.27 kg.
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