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Blababa [14]
3 years ago
12

What is the weight of a person with a massa of 80kg

Physics
1 answer:
Archy [21]3 years ago
5 0
784 Newtons or 176.37 lbs
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How are the climates of coastal regions affected by the specific heat capacity of water?
Katen [24]

Answer:

The specific heat capacity can be defined as the amount of heat required to raise the temperature of 1 unit of mass by 1 unit temperature. The specific heat capacity of water is 4.186 joule/gram °C which is higher than common substances. The land has lower specific heat capacity. Thus, the land gets hot quickly than water.

This results in warming up air near the land which creates a difference in pressure across the coastal region. Sea breeze blows from sea towards landmass. Opposite happens at night, when water is still warm and land gets cooled down quickly. Then land breeze blows  from landmass towards the sea. This breeze maintains a moderate temperature and windy and humid weather in the coastal regions.

5 0
4 years ago
For a photoelectric tube, calculate the voltage which will be just sufficient to stop electrons emitted by the sodium photo-plat
Talja [164]

Answer:

1.11 V

Explanation:

Given that the Einstein photoelectric equation states that;

KE = E - Wo

E = energy of incident photon

Wo= work function of the metal

E = hf = 6.64 x 10-34 * 6 x 1014

E = 39.84 * 10^-20 J or 3.98  * 10^-19 J

KE = 3.98  * 10^-19 J - 2.2 x 10-19J

KE = 1.78 *  10^-19J

We convert this value of KE to electron volts

KE = 1.78 *  10^-19J/1.6 x 10-19C

KE = 1.11 eV

Hence; 1.11 V will be just sufficient to stop electrons emitted by the sodium photo-plate reaching the collector plate.

8 0
3 years ago
If 7 ounces of a sports drink contains 110 milligrams of sodium, what is the total number of milligrams of sodium in 20 ounces o
ehidna [41]

so you just take 110 divided by 7 and then you get the answer and times tthat by 20 and you get you answer which is 314.28 milligrams of sodium in 20 ounces of the sports drink.

7 0
3 years ago
A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w
Scilla [17]

Answer:

v_f = 20 m/s

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2

now for pure rolling condition we will have

v = R\omega

also we have

I = mR^2

now we will have

KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}

KE = mv^2

now by work energy theorem we can say

W = KE_f - KE_i

842 J = mv_f^2 - mv_i^2

842 = 3(v_f^2) - 3\times 11^2

now solve for final speed

v_f = 20 m/s

3 0
3 years ago
A net torque of magnitude 600 Nm is ex-
vlabodo [156]
A net torque of magnitude is 600

5 0
3 years ago
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