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Marysya12 [62]
2 years ago
9

When the mass of an object increases, the forcé of gravity​

Physics
2 answers:
diamong [38]2 years ago
7 0

Answer:

increace

Explanation:

they are both going up

slava [35]2 years ago
3 0
Hi I hope you have a good day today
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Number of valence electrons for calcium?
zimovet [89]
Calcium has 2 valence electrons
7 0
3 years ago
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A charged particle of mass m = 5.00g and charge q = -70.0μC moves horizontally to the right at a constant velocity of v = 30.0 k
pychu [463]

Answer:

The magnitude of the force, B = 5 Tesla, Up (North) direction

Explanation:

Magnetic force F= Eq where Electric field, E = 750 NC

and charge, q = -70 μC = -7 ×10^{-5}C

F = 750 ×  -7 ×10^{-5}

F = 0.0525

But F = qvB; B = \frac{F}{qv}

where B is the magnetic field

= 0.0525 ÷ ( -7 ×10^{-5} × 30)

B = 5.0 Teslas

The force on a negative charge is in exactly the opposite direction to that on a positive charge.

Hence the direction of the charge is up (North).

8 0
3 years ago
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Estimate the power you produce in running up a flight of stairs. Give your answer in horsepower (1 hp = 746 W). Suppose you clim
GarryVolchara [31]
The first thing you should do is calculate the work done when climbing the stairs. This work by definition will be given by:
 W = F * d
 W = (m * g) * (d)
 W = ((71) * (9.8)) * (3) = 2087.4J
 Then, you can calculate the power that in this case is given by
 P = W / t
 P = (2087.4) / (10) = 208.74W
 To have the result in HP we use the fact that 1HP = 746W
 P = (208.74) / (746)
 P = 0.28 HP
 answer
 the power you produce in running up a flight of stairs is 0.28 HP
3 0
3 years ago
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Would someone plz help a shister outttt
tiny-mole [99]

Yaaaas, do you watch James Charles!?

8 0
3 years ago
A 0.54 kg bullfrog is sitting at rest on a level log. How large is the normal force of the log on the bullfrog?
RUDIKE [14]

Answer:

<h2>5.3N</h2>

Explanation:

Step one:

given data

mass of bullfrog= 0.54kg

Step two:

Required

The force F in Newton

From newton first law

F=mg

The acceleration due to gravity acting on the frog is 9.81m/s^2

hence

F=0.54*9.81

F=5.3N

The normal force is 5.3N

7 0
3 years ago
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