All categories

3 years ago

What is the force weight of a jaguar who jumps 3 meters to a tree branch with 2670 J of work?

Physics

1 answer:

fgiga [73]3 years ago

7 0

Answer:

$we \: know \: energy \: = \: force \: \times distance \\ e = f \times d \\ so \: f \: = \frac{e}{d} \\ so \: th \: force \: here \: = (\frac{2670}{3}) newton \\ = 890newton$

Hope it helps

You might be interested in

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is:

$a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)$

$a_{t} = -1.075\, \frac{m}{s^{2}}$

The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) If we know that $\omega_{o} = 104.720\,\frac{rad}{s}$ and $\omega = 26.180\,\frac{rad}{s}$ , then the time needed is:

$26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}$

$5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}$

$t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}$

$t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }$

$t \approx 3.963\,s$

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

8 0

3 years ago

A uniform bar has two small balls glued to its ends. The bar has length L and mass M, while the balls each have mass m and can b

vazorg [7]

Answer:

Explanation:

Length of bar = L

mass of bar = M

mass of each ball = m

Moment of inertia of the bar about its centre perpendicular to its plane is

$I_{1}=\frac{ML^{2}}{12}$

Moment of inertia of the two small balls about the centre of the bar perpendicular to its plane is

$I_{2}=2\times m\times \frac{L^{2}}{4}$

$I_{2}=\frac{mL^{2}}{2}$

Total moment of inertia of the system about the centre of the bar perpendicular to its plane is

I = I1 + I2

$I=\frac{ML^{2}}{12}+\frac{mL^{2}}{2}$

$I=\frac{(M +6m)L^{2}}{12}$

8 0

3 years ago

How does Pascal's principle describe pressure throughout a fluid?

jarptica [38.1K]

Answer:

Uniform

Explanation:

The Pascal's principle states that a change in pressure applied to an enclosed fluid is transmitted unchanged to all parts of the fluid and to the container's wall.

This implies that there is no change in magnitude of pressure at every point of the fluid and the walls of the container. Hence you can say that pressure is equal in all directions at any point of the fluid.

8 0

3 years ago

A moving walkway has a speed of 0.9 m/s to the east. A stationary observer sees a man walking on the walkway with a velocity of

Leona [35]

B or .2 East is the correct answer

3 0

3 years ago

Read 2 more answers

The density of a mineral is determined by its mass and

My name is Ann [436]

Answer: specific gravity

8 0

3 years ago