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4 years ago

13

a man with a mass of 80 kg is standing on the edge of a table 1.5 m above the ground. what is the man's gravitational potential

energy?

1 answer:

Viktor [21]4 years ago

4 0

80 kg × 1.5 m × 10 m/sec²

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a solid metal sphere of radius 3.00m carries a total charge of -5.50. what is the magnitude of the electric field at a distance

aivan3 [116]

Answer:

(a) Electric field at 0.250 m is zero.

(b) Electric field at 2.90 m is zero.

(c) Electric field at 3.10 m is - 5.15 x 10³ V/m.

(d) Electric field at 8.00 m is - 0.77 x 10³ V/m.

Explanation:

Let Q and R are the charge and radius of the solid metal sphere. The solid metal sphere behave as conductor, so total charge Q is on the surface of the sphere.

Electric field inside and outside the metal sphere is :

E = 0 for r ≤ R ( inside )

= $\frac{KQ}{r^{2} }$ for r > R ( outside )

Here K is electric constant and r is the distance from the center of the metal sphere.

(a) Electric field at 0.250 m is zero as r < R i.e. 0.250 m < 3 m from the above equation.

(b) Electric field at 2.90 m is zero as r < R i.e. 2.90 m < 3 m from the above equation.

(c) Electric field at 3.10 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 3.10 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{3.10^{2} }$

E = - 5.15 x 10³ V/m

(d) Electric field at 8.00 m is given by the relation as r > R :

E = $\frac{KQ}{r^{2} }$

Substitute 9 x 10⁹ N m²/C² for K, -5.50 μC for Q and 8.00 m for r in the above equation.

E = - $\frac{9\times10^{9}\times5.50\times10^{-6} }{8^{2} }$

E = - 0.77 x 10³ V/m

8 0

3 years ago

How often must you check the temperature of food that is being held with temperature control?.

loris [4]

You need to check the temperature of food being stored in a temperature-controlled environment every four hours. The process of changing a space's temperature is called temperature control.

Cooking food alone may not be enough to avoid food poisoning, though, if the bacteria in food are allowed to grow to large numbers. When the temperature is between 5°C and 63°C, bacteria can grow. The risk zone is the range between 5°C and 63°C.

Temperature control is a process where the passage of heat energy into or out of a space or substance is adjusted to achieve the desired temperature. This process involves measuring or otherwise detecting changes in the temperature of the space (and all of the objects contained therein) or of the substance.

Learn more about temperature here

brainly.com/question/11244611

#SPJ4

6 0

1 year ago

Susan's 10.0kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30? above the f

elena-s [515]

Answer:

The speed after being pulled is 2.4123m/s

Explanation:

The work realize by the tension and the friction is equal to the change in the kinetic energy, so:

$W_T+W_F=K_f-K_i$ (1)

Where:

$W_T=T*x*cos(0)=32N*3.2m*cos(30)=88.6810J\\W_F=F_r*x*cos(180)=-0.190*mg*x =-0.190*10kg*9.8m/s^{2}*3.2m=59.584J\\ K_i=0\\K_f=\frac{1}{2}*m*v_f^{2}=5v_f^{2}$

Because the work made by any force is equal to the multiplication of the force, the displacement and the cosine of the angle between them.

Additionally, the kinetic energy is equal to $\frac{1}{2}mv^{2}$ , so if the initial velocity $v_i$ is equal to zero, the initial kinetic energy $K_i$ is equal to zero.

Then, replacing the values on the equation and solving for $v_f$ , we get:

$W_T+W_F=K_f-K_i\\88.6810-59.5840=5v_f^{2}\\29.097=5v_f^{2}$

$\frac{29.097}{5}=v_f^{2}\\\sqrt{5.8194}=v_f\\2.4123=v_f$

So, the speed after being pulled 3.2m is 2.4123 m/s

8 0

3 years ago

The number of magnetic field force lines passing through the given surface determines: 1. magnetic flux 2. magnetic induction 3.

Assoli18 [71]

A magnetic flux would be the correct answer

8 0

4 years ago

During a race, a runner runs at a speed of 6 m/s. 2 seconds later, she is running at a speed of 10 m/s. What is the runner’s acc

Lapatulllka [165]

Let's calculate the average acceleration. It is the rate of changing speeds. Hence, we need to calculate the difference of speeds. 10-6=4 m/s. The rate is now $\frac{4m/s}{2s} =2m/s^2$ .
In general, the formula for the mean acceleration between two times 1 and 2 is given by:
$\frac{u_2-u_1}{T}$ where v1 and v2 are the speeds at the respective points and T is the time interval between them.

5 0

3 years ago