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marusya05 [52]
3 years ago
10

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

Physics
1 answer:
NISA [10]3 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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MathPhys Help pls Tysm
HACTEHA [7]

Answer:

8.75

Explanation:

First, find the force of friction.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.9 m/s)² = F (1.4 m)

F = 11.7 N

Next, find the distance at the new velocity.

Kinetic energy = work done by friction

½ mv² = Fd

½ (3.9 kg) (2.5 × 2.9 m/s)² = (11.7 N) d

d = 8.75 m

3 0
3 years ago
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What is the function of layer of air trapped under the hovercraft​
aev [14]

Answer:b

Explanation:

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3 years ago
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d
zvonat [6]

Answer:

<u />D_l=d<u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

V_{e1}=V_{e2}

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

D=\frac{at^2}{2}

Where

Acceleration is given as

a=\frac{V_o}{2d}

And

Time

T=\frac{d}{v_0}

Therefore horizontal displacement towards the left is

D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}

<u />D_l=d<u />

5 0
2 years ago
4. What is malleability?
jok3333 [9.3K]

Answer: B) The ability of steel to be shaped

4 0
3 years ago
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Electrons are accelerated in the picture tube of a television through potential difference of 8.00 * 10 ^ 3 V. (Use the values q
maria [59]

Answer:

\lambda=1.37\times 10^{-11}\ m

Explanation:

Give that,

The potential difference of the electrons, V=8\times 10^{3}\ V

We need to find the wavelength of the electrons.

Using the conservation of energy,

2meV=\dfrac{h^2}{\lambda^2}\\\\\lambda=\sqrt{\dfrac{h^2}{2meV}}

Put all the values,

\lambda=\sqrt{\dfrac{(6.63\times 10^{-34})^2}{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 8\times 10^3}}\\\\\lambda=1.37\times 10^{-11}\ m

So, the wavelength of the electrons is 1.37\times 10^{-11}\ m.

7 0
2 years ago
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