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3 years ago

What affects the vertical (y-component) of a projectile?

Physics

2 answers:

Shtirlitz [24]3 years ago

8 0

A. the force of gravity

34kurt3 years ago

5 0

Answer:

Gravity

EXPLANATION:

In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity is not constant, but changes. The y component of the acceleration ay is the downward acceleration due to gravity.

<u><em>PLS MARK BRAINLIEST</em></u>

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What is the correct answer?

pentagon [3]

Answer:

B) x^2+6x+8

Explanation:

x-4 | x^3+2x^2-16x-32

- x^3-4x^2 <-- (x-4)(x^2)

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6x^2-16x-32

- 6x^2-24x <-- (x-4)(6x)

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8x-32

- 8x-32 <- (x-4)(8)

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0 | x^2+6x+8

This means the answer is B) x^2+6x+8

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2 years ago

When jogging outside you accidentaly bumb into a curb. Your feet stop but your body continues to movr foward and you end up on t

sattari [20]

B.) Newtons first law of motion

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3 years ago

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A hot cube of iron was heated up using 1500 J of thermal energy and was placed in a beaker of water. Before it was heated, the i

Mariana [72]

Answer:

451.13 J/kg.°C

Explanation:

Applying,

Q = cm(t₂-t₁)............... Equation 1

Where Q = Heat, c = specific heat capacity of iron, m = mass of iron, t₂= Final temperature, t₁ = initial temperature.

Make c the subject of the equation

c = Q/m(t₂-t₁).............. Equation 2

From the question,

Given: Q = 1500 J, m = 133 g = 0.113 kg, t₁ = 20 °C, t₂ = 45 °C

Substitute these values into equation 2

c = 1500/[0.133(45-20)]

c = 1500/(0.133×25)

c = 1500/3.325

c = 451.13 J/kg.°C

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3 years ago

The purpose of the little Albert experiment?

HACTEHA [7]

Answer:

The aim of Watson and Rayner was to condition a phobia in an emotionally stable child.

Explanation:

Does this help?

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3 years ago

A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7

aleksandrvk [35]

(a) 1440.5 Hz

The general formula for the Doppler effect is

$f'=(\frac{v+v_r}{v+v_s})f$

where

f is the original frequency

f is the apparent frequency

$v$ is the velocity of the wave

$v_r$ is the velocity of the receiver (positive if the receiver is moving towards the source, negative otherwise)

$v_s$ is the velocity of the source (positive if the source is moving away from the receiver, negative otherwise)

Here we have

f = 1110 Hz

v = 334 m/s

In the reflector frame (= on surface B), we have also

$v_s = v_A = -28.7 m/s$ (surface A is the source, which is moving towards the receiver)

$v_r = +62.2 m/s$ (surface B is the receiver, which is moving towards the source)

So, the frequency observed in the reflector frame is

$f'=(\frac{334 m/s+62.2 m/s}{334 m/s-28.7 m/s})1110 Hz=1440.5 Hz$

(b) 0.232 m

The wavelength of a wave is given by

$\lambda=\frac{v}{f}$

where

v is the speed of the wave

f is the frequency

In the reflector frame,

f = 1440.5 Hz

So the wavelength is

$\lambda=\frac{334 m/s}{1440.5 Hz}=0.232 m$

(c) 1481.2 Hz

Again, we can use the same formula

$f'=(\frac{v+v_r}{v+v_s})f$

In the source frame (= on surface A), we have

$v_s = v_B = -62.2 m/s$ (surface B is now the source, since it reflects the wave, and it is moving towards the receiver)

$v_r = +28.7 m/s$ (surface A is now the receiver, which is moving towards the source)

So, the frequency observed in the source frame is

$f'=(\frac{334 m/s+28.7 m/s}{334 m/s-62.2 m/s})1110 Hz=1481.2 Hz$

(d) 0.225 m

The wavelength of the wave is given by

$\lambda=\frac{v}{f}$

where in this case we have

v = 334 m/s

f = 1481.2 Hz is the apparent in the source frame

So the wavelength is

$\lambda=\frac{334 m/s}{1481.2 Hz}=0.225 m$

8 0

3 years ago