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diamong [38]
4 years ago
8

All Physicsts over here plz help in these questions!!!!!!!!!

Physics
2 answers:
mrs_skeptik [129]4 years ago
6 0
Hello!

First one we can use that PE=mgh so we have

4.37*10^5J/(9.12*10^3kg*9.80m/s^2)= 4.89m

Second one we can use Newton’s Second Law
F=ma and in this case F=mg so we have

g= 3.28*10^-2N/6*10^-3kg = 5.47m/s^2

Hope this helps. Any questions please ask. Thank you.
Vinil7 [7]4 years ago
6 0

1. Answer: Height = 4.8 m

Gravitational potential energy, P.E. = mgh

where, m is mass of the body, g is the acceleration due to gravity and h is the height.

It is given that, location is on Earth.

P.E. =4.37\times 10^5 J\\ m=9.12 \times 10^3 kg\\ g=9.8 m/s^2

\Rightarrow h=\frac{P.E.}{mg}=\frac{4.37\times 10^5 J}{9.12 \times 10^3 kg \times 9.8 m/s^2}=4.8 m

2. Answer: 5.46m/s^2

Weight is a force which an object exerts due to its mass under gravity.

weight = mass × acceleration due to gravity

⇒ acceleration due to gravity = weight/ mass

It is given that the weight, w= 3.28\times 10^{-2} N

mass of the object, m=6\times 10^{-3} kg

\Rightarrow a=\frac{w}{m}=\frac{3.28\times 10^{-2} N}{6\times 10^{-3} kg}=5.46 m/s^2


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What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a
Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure =  P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0
4 years ago
Ramesh announced in class: ''Yesterday I had fever and my body temperature was 100 degrees.'' Ravi said: ''We learnt in the last
pogonyaev

Answer:

D. Ramesh and Ravi are correct, but they are using different measurement scales.

\Huge{\underline{\textrm{Explanation}}}Explanation

Here, Ravi says that his body temperature is 100 degrees, but does not mention that whether it is 100 degrees Celsius or 100 degrees Fahrenheit. When the temperature of a human body is more than 100.4 degree Fahrenheit (38°C), or near to it, the person is considered to have fever.

The boiling point of water is 100 degrees Celsius and not 100 degrees Fahrenheit.

Thus, they both are using different measurement scales.

7 0
3 years ago
During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi
geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

a_t = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

\omega_i = Initial angular velocity = 95 rad/s

\theta = Angle of rotation

\omega_f = Final angular velocity

t = Time taken

Angular acceleration is given by

\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2

The angular acceleration is -24.28571 rad/s²

\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev

The number of revolutions is 29.57239

\omega_f=\omega_i+\alpha t\\\Rightarrow t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-95}{-24.28571}\\\Rightarrow t=3.91176\ s

The time it takes for the car to stop is 3.91176 seconds

Linear distance

s=r\theta\\\Rightarrow s=0.28\times 185.80885\\\Rightarrow s=52.026478\ m

The distance the car travels is 52.026478 m

8 0
3 years ago
Astronomers use many different pieces of equipment to study space: optical telescopes on Earth, space telescopes, and space prob
Kobotan [32]
Then it would be C if D has been eliminated
8 0
3 years ago
Read 2 more answers
The current supplied by a battery as a function of time is I(t) = (0.88 A) e^(-t*6 hr). What is the total number of electrons tr
Degger [83]

Answer:

e. 1.2 x 10²³

Explanation:

According to the problem, The current equation is given by:

I(t)=0.88e^{-t/6\times3600s}

Here time is in seconds.

Consider at t=0 s the current starts to flow due to battery and the current stops when the time t tends to infinite.

The relation between current and number of charge carriers is:

q=\int\limits {I} \, dt

Here the limits of integration is from 0 to infinite. So,

q=\int\limits {0.88e^{-t/6\times3600s}}\, dt

q=0.88\times(-6\times3600)(0-1)

q = 1.90 x 10⁴ C

Consider N be the total number of charge carriers. So,

q = N e

Here e is electronic charge and its value is 1.69 x 10⁻¹⁹ C.

N = q/e

Substitute the suitable values in the above equation.

N= \frac{1.9\times10^{4} }{1.69\times10^{-19}}

N = 1.2 x 10²³

8 0
3 years ago
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