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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

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A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a

Elza [17]

Answer:

$\dfrac{dz}{dt}=0.65\ ft/s$

Explanation:

Given that

x= 150 ft

$\dfrac{dy}{dt}= 7\ ft/s$

y= 14 ft

From the diagram

$z^2=x^2+y^2$

When ,x= 150 ft and y= 14 ft

$z^2=150^2+14^2$

$z=\sqrt{150^2+15^2}$

z=150.74 ft

$z^2=x^2+y^2$

By differentiating with respect to time t

$2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}$

$z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$

Here x is constant that is why

$\dfrac{dx}{dt}=0$

$z\dfrac{dz}{dt}= y\dfrac{dy}{dt}$

Now by putting the values in the above equation we get

$150.74\times \dfrac{dz}{dt}=14\times 7$

$\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s$

$\dfrac{dz}{dt}=0.65\ ft/s$

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0

3 years ago

Suppose I have a vector that is 7 units long and that makes an angle of +30 degrees from the positive x-axis. I want to add to t

Vinil7 [7]

Answer:

sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

Explanation:

We have given first vector which has length of 7 units and makes an angle of 30° with positive x-axis

So x component of the vector $=7cos30^{\circ}=7\times 0.866=6.06$

y component of the vector $=7sin30^{\circ}=7\times 0.5=3.5$

So vector will be 6.06i+3.5j

Now other vector of length of 7 units and makes an angle of 120° with positive x-axis

So x component of vector $=7cos120^{\circ}=7\times -0.5=-3.5i$

y component of the vector $=7sin120^{\circ}=7\times 0.866=6.06j$

Now sum of these two vectors is 6.06i+3.5j-3.5i+6.06j = 2.56i+9.56j

5 0

4 years ago

Ohm’s law describes the relationship between which quantities?

VladimirAG [237]

I think its between b or c

5 0

4 years ago

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The near point of an eye is 48.5 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan

Masteriza [31]

Answer:

The focal length of the lens should be -51.5 cm (a concave lens).

Explanation:

The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.

We use the lens formula

$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

where f = focal length, u = object distance and v = image distance.

In this case, u = 48.5 cm and v = -25.0 cm.

v is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)

$\dfrac{1}{f} = \dfrac{1}{48.5} + \dfrac{1}{-25.0} = -\dfrac{23.5}{1212.5}$

$f = -51.5$

The negative sign indicates the lens is concave.

3 0

3 years ago

Which is not a way to accelerate an object?

densk [106]

I would definitely think its B....

3 0

3 years ago

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