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3 years ago

What type of front occurs when cold air and warm air are next to each other but are at a standstill

Physics

1 answer:

Korvikt [17]3 years ago

5 0

Answer:

A stationary front.

(Hope this helped! c:)

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6.6 kg block initially at rest is pulled to theright along a horizontal, frictionless surfaceby a constant, horizontal force of

neonofarm [45]

Answer:

v = 3.04 m/s

Explanation:

given,

mass of the block, M = 6.6 Kg

horizontal force, F = 12.2 N

distance, L = 2.5 m

initial speed = 0 m/s

speed of the block,v = ?

we now

Work done is equal to change in Kinetic energy.

Work done = Force x displacement

W = Δ K E

Δ K E = Force x displacement

$\dfrac{1}{2}mv^2 - \dfrac{1}{2}mu^2= F .s$

$\dfrac{1}{2}\times 6.6 \times v^2 - 0= 12.2\times 2.5$

3.3 v² = 30.5

v² = 9.242

v = 3.04 m/s

speed of the block is equal to 3.04 m/s

4 0

2 years ago

Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?

Ede4ka [16]

Answer:

W=f×d

w=45×27

w=1215 j.............

3 0

3 years ago

You throw a glove straight upward to celebrate a victory. Its initial kinetic energy is K and it reaches a maximum height h. Wha

Rasek [7]

Answer:

K/2

Explanation:

The law of conservation of mechanical energy states that the sum of the kinetic and potential energies is a constant at any point.

At maximum height, the glove has purely potential energy but at the bottom, it has purely kinetic energy.

The potential energy at the top = kinetic energy at the bottom. The potential energy is given by

$PE = mgh$

At half height, this potential energy is

$PE = \frac{1}{2}mgh$

At this height, PE + KE = Constant = KE at bottom or PE at maximum height.

$mgh = \frac{1}{2}mgh +KE$

$KE = \frac{1}{2}mgh = K/2$

5 0

3 years ago

I will pick the most Brainliest but... i need u to tell me how too

katovenus [111]

Answer:

I think it’s a

Explanation:

8 0

3 years ago

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Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each

jarptica [38.1K]

Answer with Explanation:

We are given that

$I_1=29 A$

$I_2=78 A$

$d=38 cm=\frac{38}{100}=0.38 m$

$1 m=100 cm$

a.Length of segment,l=20 m

Magnetic force ,F= $\frac{2\mu_0I_1I_2 l}{4\pi d}$

$\frac{\mu_0}{4\pi}=10^{-7}$

Substitute the values

$F=\frac{10^{-7}\times 29\times 78\times 20}{0.38}=0.0119 N$

Hence, the magnetic force exert by each segment on the other=0.0119 N

b.We know that when current carrying in the wires are in same direction then the force will attract to each other.

Hence, the force will be attractive.

4 0

2 years ago

Read 2 more answers