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Finger [1]
3 years ago
11

E)

Physics
1 answer:
guapka [62]3 years ago
7 0

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on  her one foot. The force acting per unit area is called pressure. It can be given by :

P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa

So, the pressure is 32666.66 Pa.

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A 500.-kg roller coaster car starts from rest at the top of a 60.0-meter hill.
Paraphin [41]

1.47x10^5 Joules  
The gravitational potential energy will be the mass of the object, multiplied by the height upon which it can drop, multiplied by the local gravitational acceleration. And since it started at the top of a 60.0 meter hill, halfway will be at 30.0 meters. So  
500 kg * 30.0 m * 9.8 m/s^2 = 147000 kg*m^2/s^ = 147000 Joules.  
Using scientific notation and 3 significant figures gives 1.47x10^5 Joules.
8 0
3 years ago
The potential difference between points A and B in an electric
Flauer [41]

Answer:

i hope  the answear is D becuase  went over this long time ago when i was like you

Explanation:

3 0
3 years ago
What speed must a 600 kg car have in order to have the same momentum as a 1200 kg truck traveling at a velocity of 10 m/s to the
yKpoI14uk [10]

Answer:

20 m/s

Explanation:

4 0
3 years ago
A 20-turn coil of area 0.32 m2 is placed in a uniform magnetic field of 0.055 T so that the perpendicular to the plane of the co
makvit [3.9K]

Answer:

1.5 * 10^-2 Tm^2

Explanation:

Electric Flux = B.A cos(theta)

B = 0.055 T

A = 0.32 m^2

theta = 30

Electric Flux = (0.055 T).(0.32 m^2).Cos(30) = 0.0152 = 1.5 * 10^-2 Tm^2

5 0
2 years ago
A beam of light, with a wavelength of 4170 Å, is shined on sodium, which has a work function (binding energy) of 4.41 × 10 –19 J
Lyrx [107]

Explanation:

Given that,

Wavelength of the light, \lambda=4170\ A=4170\times 10^{-10}\ m

Work function of sodium, W_o=4.41\times 10^{-19}\ J

The kinetic energy of the ejected electron in terms of work function is given by :

KE=h\dfrac{c}{\lambda}-W_o\\\\KE=6.63\times 10^{-34}\times \dfrac{3\times 10^8}{4170\times 10^{-10}}-4.41\times 10^{-19}\\\\KE=3.59\times 10^{-20}\ J

The formula of kinetic energy is given by :

KE=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2KE}{m}} \\\\v=\sqrt{\dfrac{2\times 3.59\times 10^{-20}}{9.1\times 10^{-31}}} \\\\v=2.8\times 10^5\ m/s

Hence, this is the required solution.

7 0
3 years ago
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