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Colt1911 [192]
3 years ago
6

Please tell the correct answer​

Physics
1 answer:
alukav5142 [94]3 years ago
8 0

Answer:

The answer is C she didn't mark C so its your best best of trying.Good Luck!

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What does MA mean for you in real life?
andrezito [222]
To me "Ma" mean in real life is either mama or Massachusttes.
8 0
3 years ago
What force does a trampoline have to apply to a 45.0-kg gymnast to accelerate her straight up at 7.50 m/s^2? (a) 104N (b) 338 N
Brilliant_brown [7]

Answer:

b) 338 N

Explanation: let m be the mass of the gymnast and a be the acceleration of the gymnast.

the force required to accelerate the gymnast is given by:

F = m×a

  = (45.0)×(7.50)

  = 337.5 N

Therefore, the force a trampoline has to apply is 138 N.

6 0
3 years ago
When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g
mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=mc_p dT

Where m is the mass c_p = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

T_1 =25°C

T_2 we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25-T_2)

25-T_2 =22

T_2=3°C

7 0
3 years ago
Read 2 more answers
A uniform meterstick of mass 0.20 kg is pivoted at the 40 cm mark. where should one hang a mass of 0.50 kg to balance the stick?
Tcecarenko [31]
The weight of the meterstick is:
W=mg=0.20 kg \cdot 9.81 m/s^2 = 1.97 N
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance 
d_1 = 0.50 m - 0.40 m=0.10 m
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
M_w = W d_1 = (1.97 N)(0.10 m)=0.20 Nm

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
(mg) d_2 = 0.20 Nm
from which we find the value of d2:
d_2 =  \frac{0.20 Nm}{mg}= \frac{0.20 Nm}{(0.5 kg)(9.81 m/s^2)}=0.04 m

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
4 0
3 years ago
If the magnitude of a charge is half as much as another charge, but the force experienced is the same, then the electric field s
Kazeer [188]

Answer:

the electric field strength of this charge is two times the strength of the other charge

Explanation:

Using the relationship between electric field and the charge, which is inversely proportionality. Let the the magnitude of the first charge be Q and the respective electric field be E. It implies that;

E1/E2 = Q2/Q1

E2 = E1 x Q1/Q2

      = E x Q/ (Q/2)

       = 2E

8 0
3 years ago
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