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2 years ago

A bird flies north with a force of 11N, the air resistance is 2N. How do i do a diagram of this?

Physics

1 answer:

Novay_Z [31]2 years ago

7 0

This is a diagram from above, the air resistance is oppose to movement, the bird is moving forward given its force is bigger.

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Use the diagram below modeling a football kicked from a horizontal surface B

djyliett [7]

B is the correct one

5 0

2 years ago

Read 2 more answers

A seagull flying horizontally at 8.00m/s carries a clam with a mass of 300g in its beak. Calculate the total mechanical energy o

Stells [14]

Answer:

9.6J+88.2J=97.8J

Explanation:

Here the velocity of the seagull is given,mass is given and its height.

We have to find its mechanical energy my friend.

Mechanical energy=kinetic energy + potential energy.

First we will find kinetic energy.

For calculating kinetic energy we need mass and velocity,which are given here.

So, Ek=

$1 \div 2mv {?}^{2}$

So by substituting the values we get 9.6J.

Now we find the potential energy which is mgh.

By substituting the values we get 88.2J.

Then we add both of those and get 97.8J

I hope this satisfies you and make sure you contact me if it doesn't

7 0

3 years ago

What is the exact meaning of net force?

ZanzabumX [31]

In science and physics net force is the mean or overall of all the forces acting on an object.

4 0

2 years ago

A square-based shipping crate is being designed that must contain a volume of 16 ft3 . The material that is used for the base an

Vlada [557]

Answer:

Explanation:

Given

volume $V=16 ft^3$

Suppose base is square with side L

height of crate is h

Volume $V=L^2\times h$

$16=L^2\times h$

Cost of top and bottom area $c_1=3L^2$

Cost of Side area $c_2=4Lh\times 2=8Lh=8L\times \frac{16}{L^2}=\frac{128}{L}$

Total Cost $C=c_1+c_2$

Total Cost $C=3L^2+\frac{128}{L}$

Differentiate C w.r.t Length

$\frac{dC}{dL}=6L-\frac{128}{L^2}$

$L^3=\frac{128}{6}$

$L=2.75 ft$

$h=\frac{16}{2.75^2}=11.46 ft$

Dimensions are $L\times L\times h=2.75\times 2.75\times 11.46$

6 0

2 years ago

A 0.20-kg mass is oscillating on a spring over a horizontal frictionless surface. When it is at a displacement of 2.6 cm for equ

valentinak56 [21]

Explanation:

The given data is as follows.

mass = 0.20 kg

displacement = 2.6 cm

Kinetic energy = 1.4 J

Spring potential energy = 2.2 J

Now, we will calculate the total energy present present as follows.

Total energy = Kinetic energy + spring potential energy

= 1.4 J + 2.2 J

= 3.6 Joules

As maximum kinetic energy of the object will be equal to the total energy.

So, K.E = Total energy

= 3.6 J

Also, we know that

K.E = $\frac{1}{2}mv^{2}_{m}$

or, v = $\sqrt{\frac{2K.E}{m}}$

= $\sqrt{2 \times 3.6 J}{0.2 kg}$

= $\sqrt{36}$

= 6 m/s

thus, we can conclude that maximum speed of the mass during its oscillation is 6 m/s.

4 0

3 years ago