All categories

3 years ago

if you have a kinetic energy of 1470 J, and you are 60kg mass and 0 m above the ground, what is you velocity?

Physics

2 answers:

laiz [17]3 years ago

8 0

Answer:

The 39.

Explanation:

Anni [7]3 years ago

6 0

KE = 1/2 · m · v²

1470 = (1/2 · 60) · (v²)

v² = 1470 / 30

v² = 49

v = √49

(Note: Kinetic energy has nothing to do with height above ground.)

You might be interested in

What travels in a wave from one location to another is _________.

Alina [70]

Answer:

Energy

Explanation:

A wave is a kind of disturbance which passes through a medium and during this process, energy is transferred from one point to another without causing any permanent displacement to the medium itself.

for example if a stone is dropped inside water, circular ripples are observed and energy moves from one point to another .

5 0

3 years ago

A person trapped outside during a thunderstorm should

Katyanochek1 [597]

Answer:

should stay away from trees, water, and tall objects.

6 0

3 years ago

Read 2 more answers

Can someone help me with these

Lostsunrise [7]

Answer:

what is this question

Explanation:

sorry sorry

8 0

3 years ago

An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the ce

Mamont248 [21]

Answer:

Explanation:

To find the angular velocity of the tank at which the bottom of the tank is exposed

From the information given:

At rest, the initial volume of the tank is:

$V_i = \pi R^2 h_i --- (1)$

where;

height h which is the height for the free surface in a rotating tank is expressed as:

$h = \dfrac{\omega^2 r^2}{2g} + C$

at the bottom surface of the tank;

r = 0, h = 0

∴

$h = \dfrac{\omega^2 r^2}{2g} + C$

0 = 0 + C

C = 0

Thus; the free surface height in a rotating tank is:

$h=\dfrac{\omega^2 r^2}{2g} --- (2)$

Now; the volume of the water when the tank is rotating is:

dV = 2π × r × h × dr

Taking the integral on both sides;

$\int \limits ^{V_f}_{0} \ dV = \int \limits ^R_0 \times 2 \pi \times r \times h \ dr$

replacing the value of h in equation (2); we have:

$V_f} = \int \limits ^R_0 \times 2 \pi \times r \times ( \dfrac{\omega ^2 r^2}{2g} ) \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \int \limits ^R_0 \ r^3 \ dr$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{r^4}{4} \Big]^R_0$

$V_f = \dfrac{ \pi \omega ^2}{g} \Big [ \dfrac{R^4}{4} \Big] --- (3)$

Since the volume of the water when it is at rest and when the angular speed rotates at an angular speed is equal.

Then $V_f = V_i$

Replacing equation (1) and (3)

$\dfrac{\pi \omega^2}{g}( \dfrac{R^4}{4}) = \pi R^2 h_i$

$\omega^2 = \dfrac{4g \times h_i }{R^2}$

$\omega =\sqrt{ \dfrac{4g \times h_i }{R^2}}$

$\omega = \sqrt{\dfrac{4 \times 9.81 \ m/s^2 \times 0.7 \ m}{(0.5)^2} }$

$\omega = \sqrt{109.87 }$

$\mathbf{\omega = 10.48 \ rad/s}$

Finally, the angular velocity of the tank at which the bottom of the tank is exposed = 10.48 rad/s

6 0

3 years ago

There is a potential difference of 1.1 V between the ends of a 10 cm long graphite rod that has a cross-sectional area of 0.90 m

Serga [27]

Explanation:

It is given that,

Potential difference between the ends of a rod, V = 1.1 V

Length of the rod, l = 10 cm = 0.1 m

Area of cross section of the rod, $A=0.9\ mm^2=9\times 10^{-7}\ m^2$

The resistivity of graphite, $\rho=7.5\times 10^{-6}\ \Omega-m$

(a) Let R is the resistance of the rod. It is given by :

$R=\rho \dfrac{l}{A}$

$R=7.5\times 10^{-6}\times \dfrac{0.1}{9\times 10^{-7}}$

$R=0.833\ \Omega$

So, the resistance of the rod is 0.833 ohms.

(b) Let I is the current flowing in the wire. It can be calculated using the Ohm's law as :

$I=\dfrac{V}{R}$

$I=\dfrac{1.1\ V}{0.833\ \Omega}$

I = 1.32 A

(c) Let E is the electric field inside the rod. The electric field in terms of potential difference is given by :

$E=\dfrac{V}{l}$

$E=\dfrac{1.1\ V}{0.1\ m}$

E = 11 V/m

Hence, this is the required solution.

8 0

3 years ago