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Komok [63]
3 years ago
5

A stone is thrown upward with an initial upward velocity of 15 m/s from a

Physics
1 answer:
tensa zangetsu [6.8K]3 years ago
7 0

Answer

for every meters it will go up 15 so if it was 2 secoonds it woudl be 30 and if it was 3 seconds it would be 45 meters

Explanation:

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A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra
Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

E=0

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}

E\propto\dfrac{1}{z^2}

(c). In terms of R,

We need to calculate the positive distance

If E\rightarrow E_{max}

Then, \dfrac{dE}{dz}=0

\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0

Taking only positive distance

z=\dfrac{R}{\sqrt{2}}

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}

E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}

E_{max}=15418.7\ N/C

E_{max}=1.54\times10^{4}\ N/C

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z =  ∞, The electric field due to the rod is E\propto\dfrac{1}{z^2}.

(c). The positive distance is \dfrac{R}{\sqrt{2}}

(d). The maximum magnitude of electric field is 1.54\times10^{4}\ N/C

6 0
3 years ago
A 50.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and the
Dimas [21]

Answer:

98N and 147N

Explanation:

We have the following information:

m=50kg\\\mu_s =0.4\\\mu_k = 0.2\\F=140N

We can find the static fricton force as follow,

F=\mu_s * N

Where N is the normal force (mg)

F=0.3*50*9.8\\F=147N

Static friction force at 147N is greater than the force applied hence body does not move.

F=\mu_k N = 0.2*50*9.8= 98N

3 0
3 years ago
Which of the following represents energy in its most disordered form? Group of answer choices Chemical-bond energy Electromagnet
gavmur [86]

Answer: Heat Energy

Explanation:

Heat is energy in its most disordered form. heat energy is the random jostling of molecules and is therefore not organized. As cells perform the chemical reactions that generate order within, some energy is inevitably lost in the form of heat. Because the cell is not an isolated system, the heat energy produced by the cell is quickly dispersed into the cell's surroundings where it increases the intensity of the thermal motions of nearby molecules. This increases the entropy of the cell's environment and keeps the cell from violating the second law of thermodynamics.

5 0
3 years ago
Jason walks 20 m East, turns around and 20 m West, Finally, he walks 10 rn North. This takes 20 s. what is Jason's velocity​
serious [3.7K]

Answer:

0.5 m/s north

Explanation:

Take east to be +x, west to be -x, north to be +y, and south to be -y.

His displacement in the x direction is:

x = 20 m − 20 m = 0 m

His displacement in the y direction is:

y = 10 m

His total displacement is therefore 10 m north.

His velocity is equal to displacement divided by time.

v = 10 m north / 20 s

v = 0.5 m/s north

3 0
3 years ago
Se necesita subir una carga de 500 kg (4900 N) a una altura de 1.5 m deslizándola sobre una rampa inclinada. ¿Qué longitud debe
marusya05 [52]

Answer:

4.22 m

Explanation:

Una rampa es una máquina que se utiliza para levantar un objeto con una fuerza menor a la que realmente necesitarías. Cuanto mayor sea la longitud de la rampa, menor será la magnitud de la fuerza necesaria para levantar el objeto.

Dado que:

altura de la rampa = 1.5 m, carga = 4900 N, fuerza aplicada = 1633.33 N.

La fórmula de la rampa se da como:

fuerza aplicada * longitud de la rampa = peso de la carga * altura de la rampa

1633.33 * longitud de la rampa = 4900 * 1.5

longitud de la rampa = 4900 * 1.5 / 1633.33

longitud de la rampa = 4.22 m

6 0
3 years ago
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