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4 years ago

A projectile is launched at an angel into the air its verticle acceleration is g?

Physics

1 answer:

NISA [10]4 years ago

5 0

Yes it is. That's a true statement.

But what have you got against the poor angel ?

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Which is the correct answer?

Masja [62]

The correct answer is A, 2x^3 - x^2 +3x +7

3 0

3 years ago

An electron in the n = 5 level of an h atom emits a photon of wavelength 1282.17 nm. to what energy level does the electron move

lions [1.4K]

This is an interesting (read tricky!) variation of Rydberg Eqn calculation.
Rydberg Eqn: 1/λ = R [1/n1^2 - 1/n2^2]
Where λ is the wavelength of the light; 1282.17 nm = 1282.17×10^-9 m
R is the Rydberg constant: R = 1.09737×10^7 m-1
n2 = 5 (emission)
Hence 1/(1282.17 ×10^-9) = 1.09737× 10^7 [1/n1^2 – 1/25^2]
Some rearranging and collecting up terms:
1 = (1282.17 ×10^-9) (1.09737× 10^7)[1/n2 -1/25]
1= 14.07[1/n^2 – 1/25]
1 =14.07/n^2 – (14.07/25)
14.07n^2 = 1 + 0.5628
n = √(14.07/1.5628) = 3

8 0

3 years ago

Read 2 more answers

Prompt

Mamont248 [21]

Answer:

1.bandages

2.cottons

3.scissor

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6 0

3 years ago

If an athlete leaps vertically at 4.0m/s, what maximum height does he reach?

joja [24]

Hello
This is a problem of accelerated motion, where the acceleration involved is the gravitational acceleration: $g=-9.81~m/s^2$ , and where the negative sign means it points downwards, against the direction of the motion.

Therefore, we can use the following formula to solve the problem:
$v_f^2 = v_i^2 + 2gS$
where $v_i=4~m/s$ is the initial vertical velocity of the athlete, $v_f=0$ is the vertical velocity of the athlete at the maximum height (and $v_f=0~m/s$ at maximum height of an accelerated motion) and S is the distance covered between the initial and final moment (i.e., it is the maximum height). Re-arranging the equation, we get
$S= \frac{v_f^2-v_i^2}{2g}=0.82~m$

3 0

3 years ago

Polly is pushing a box across the floor with a force of 30 N. The force of gravity is -8 N, and the normal force is 8 N. Which

wlad13 [49]

Answer:

Force of friction is (-30 N).

Explanation:

The force applied on the box across the floor is 30 N.

The force of gravity is (-8 N) and the the normal force is 8 N.

It is based on Newton's third law of motion. Newton's third law of motion states that the force acting on object 1 to object 2 is equal in magnitude of the force from object 2 to 1 but in opposite direction.

Here there is force of 30 N is applied in horizontal direction. The frictional force act in opposite direction. So, the force of friction is -30 N so that box across the floor.

5 0

3 years ago