Ask question

Ask question

All categories

tankabanditka [31]

2 years ago

10

A leopard of mass 65 kg climbs 7 m up a tall tree. Calculate how much gravitational potential energy it gains. Assume g = 10 N/k

g.

1 answer:

Kay [80]2 years ago

3 0

Mass of Leopard = 65 kg

Height = 7 m

P.E = ?

We know,

P.E = mgh

= 65 * 10 * 7=4550 J

You might be interested in

A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri

Vlada [557]

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

8 0

2 years ago

. A horizontal steel spring has a spring constant of 40.0 N/m. What force must be applied to the spring in order to compress it

ANEK [815]

Ans 4 more to be exact

Explanation:

7 0

2 years ago

Certain meteorites have been examined and found to carry samples of which molecules?

ki77a [65]

Answer:

Sugars...

Explanation:

Several meteorites have been found to carry molecules of sugars that are essential for life. These sugars include Ribose, Arabinose and Xylose. These are found in meteorites that are rich in carbon. These significant discoveries can pave way in finding the origin of life on Earth.

6 0

2 years ago

A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an

Sonbull [250]

Answer:

(a) k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

k = $\frac{M}{6} (a^{2} +b^{2} )w^{2}$

k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

8 0

2 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

2 years ago