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Nat2105 [25]
2 years ago
15

Có mấy cách đo đại lượng vật lí

Physics
1 answer:
wariber [46]2 years ago
3 0

Answer:

please write in english

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A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The
uysha [10]

Answer:

f = 692 N

Explanation:

given data:

f =800N

a =1.2 m s^{2}

m= 90 kg

from newton's second law

net force F_{net} =\sum F = F_1 +F_2 +..... = ma

therefore we have from above equationF_{net} = F -f = ma

ma =F - f

putting all value to get force of friction

1.2*90 = 800 - f

f = 692 N

8 0
3 years ago
What did you notice about the angle of incidence and the angle of reflection?
Readme [11.4K]

Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

6 0
2 years ago
When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth. If the muscle generates a force of 49.
UkoKoshka [18]

Answer:

1.1397 Nm

Explanation:

When the palmaris longus muscle in the forearm is flexed, the wrist moves back and forth.

If the muscle generates a force

F =  49.5 N and r = 2.65 cm , then the torque is equal to rF

we see that r = 2.65 cm = 0.0265 m

therefore

torque = 0.0265 x 49.5

= 1.1397 Nm

4 0
3 years ago
A bolt of lightning discharges 9.7 C in 8.9 x 10^-5 s. What is the average current during the discharge?
Anastaziya [24]

Answer: 1.089\times 10^5\ A

Explanation:

Given

Charge discharged Q=9.7\C

time taken t=8.9\times 10^{-5}\ s

Current is given as rate of change of discharge i.e.

\Rightarrow I=\dfrac{Q}{t}\\\\\Rightarrow I=\dfrac{9.7}{8.9\times 10^{-5}}\\\\\Rightarrow I=1.089\times 10^5\ A

Therefore, the average current is 1.089\times 10^5\ A

3 0
2 years ago
A non-conducting sphere of radius R = 3.0 cm carries a charge Q = 2.0 mC distributed uniformly throughout its volume. At what di
BlackZzzverrR [31]

Answer:

r =3 *\sqrt{2} = 4.24 cm

Explanation:

given data

Radius of sphere 3.0 cm

charge Q = 2.0 m C

We know that maximum electric field is given as

E_{MAX}= \frac{KQ}{r^{2}}

electric field inside the sphere can be determine by using below relation

\frac{KQ}{r^{2}}= \frac{1}{2}*\frac{KQ}{R^{2}}

r = \sqrt{2}R

r =3 *\sqrt{2} = 4.24 cm

4 0
3 years ago
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