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4 years ago

12

Two people push on the same door from opposite sides as shown they will only see the door move when

1 answer:

rusak2 [61]4 years ago

5 0

One of them overpowers the other

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What is the purpose of the Bill of Rights?

NeX [460]

Answer:

i think it is To limit the rights of individual citizens

Explanation:

4 0

3 years ago

Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a)

mafiozo [28]

(a) The work done by the bird to raise the snake on the branch is equal to the product between the weight of the snake and the height of the branch above the ground:

$W_1 = mg h$

where $m=75 g=0.075 kg$ is the mass of the snake and $h=2.5 m$ is the height of the branch with respect to the ground. Substituting numbers into the equation, we find

$W_1 = (0.075 kg)(9.81 m/s^2)(2.5 m)=1.84 J$

(b) The work the bird did to raise its own centre of mass from the ground to the branch is equal to the product between the bird's weight and the height of the branch:

$W_2 = mgh$

where $m=350 g=0.35 kg$ is the mass of the bird. Substituting numbers into the equation, we find

$W_2 = (0.35 kg)(9.81 m/s^2)(2.5 m)=8.58 J$

6 0

3 years ago

A 1250 kg car starts are rest then speeds up to 15 m/s. What is the impulse of the car?

Masteriza [31]

Answer:

B)18750

Explanation:

it because

$1250 \times 15 = 18750$

7 0

3 years ago

Read 2 more answers

A charge of 7.00 mC is placed at opposite corners corner of a square 0.100 m on a side and a charge of -7.00 mC is placed at oth

andrew-mc [135]

Answer:

4.03\times10^{7}N[/tex], 135°

Explanation:

charge, q = 7 mC = 0.007 C

charge, - q = - 7 mC = - 0.007 C

d = 0.1 m

Let the force on charge placed at C due to charge placed at D is FD.

$F_{D}=\frac{kq^{2}}{DC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FD is along C to D.

Let the force on charge placed at C due to charge placed at B is FB.

$F_{B}=\frac{kq^{2}}{BC^{2}}$

$F_{B}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1^{2}}=4.41 \times 10^{7}N$

The direction of FB is along C to B.

Let the force on charge placed at C due to charge placed at A is FA.

$F_{A}=\frac{kq^{2}}{AC^{2}}$

$F_{D}=\frac{9 \times10^{9}\times 0.007 \times 0.007}{0.1 \times\sqrt{2} \times 0.1 \times\sqrt{2}}=2.205 \times 10^{7}N$

The direction of FA is along A to C.

The net force along +X axis

$F_{x}=F_{A}Cos45-F_{D}$

$F_{x}=2.205\times10^{7}Cos45-4.41\times10^{7}=-2.85\times10^{7}N$

The net force along +Y axis

$F_{y}=F_{B}-F_{A}Sin45$

$F_{y}=4.41\times10^{7}-2.205\times10^{7}Sin45=2.85\times10^{7}N$

The resultant force is given by

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}=\sqrt{(-2.85\times10^{7})^{2}+(2.85\times10^{7})^{2}}$

$F = 4.03\times10^{7}N$

The angle from x axis is Ф

tan Ф = - 1

Ф = -45°

Angle from + X axis is 180° - 45° = 135°

5 0

3 years ago

A rock is thrown downward from a height of 207 m with an initial velocity of 5.37 m/s. Approximately how fast will it it be movi

Zolol [24]

List the known information:

$x_0=207 \text{ m}\\v_0=-5.37 \text{ m/s}\\t=3.03 \text{ s}\\a=-g=-9.8 \text{ m/s}^2$

Use the kinematic equation $v=v_0+at$ .

Plug in the given values:

$v=-5.37 \text{ m/s}+(-9.8 \text{ m/s}^2)(3.03 \text{ s})=-35.064 \text{ m/s}$

This would be 35.064 m/s downward, or 35 m/s downward with significant figures taken into account.

4 0

3 years ago