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3 years ago

15

A cannon Is pointed Upward 30 degrees and fires a cannonball with a speed of 100m/sec what is the component of the cannon balls

velocity parallel to the groud

1 answer:

Kamila [148]3 years ago

6 0

Answer:

86.6 m/s

Explanation:

vₓ = v cos θ

vₓ = 100 m/s cos 30°

vₓ = 86.6 m/s

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Explain the relationship between mass and change in velocity. You may write

Klio2033 [76]

It makes me cracked at Fortnite

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3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

4 0

3 years ago

A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

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3 years ago

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One characteristic of mass wasting processes is that they a. only operate on steep slopes. b. move materials very slowly. c. ope

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Answer:

One characteristic of mass wasting processes is that they move materials relatively short distances compared to streams. - d.

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