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2 years ago

Which name is given to the force that stretches the spring with a metal that weights it down

Physics

1 answer:

bulgar [2K]2 years ago

3 0

Answer:

gravity

Explanation:

ok what i understand from your query is that a spring balance with a metal piece attached to it is weighing it down, a spring balance is measured in newtons (f=ma), the acceleration part is gravity (-9.81 m/s^2). The metal piece is attached to the string.

hope this answers your question and please take my answer with a grain of salt and refer to some webpages or videos

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What properties of the wave define why it is found within this area of the spectrum?

MatroZZZ [7]

Answer:

Explanation:

Speed and medium are properties of a wave which is in common within an area of the spectrum of visible light. i.e;

Their medium of propagation

They both travel at the same speed ( speed of light )

The properties mentioned above are properties that define that wave is found within an area spectrum for visible light.

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2 years ago

which layer of the atmosphere is directly above the troposphere A. exosphere B. mesosphere C. troposphere or D. stratosphere

lozanna [386]

The layer above it would be D. the Stratosphere

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2 years ago

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Bob is cleaning his dishwasher with white vinegar by putting a bowl of this ____________ mixture in the top of an empty dishwash

Maurinko [17]

Answer: Baking Soda

A mixture of white vinegar and baking soda in a bowl will produce a paste that is helpful in removing stains and sanitizing a dishwasher. To meticulously clean your dishwasher, spread the mixture onto the build-up crust in the bottom basin of the washer itself then let it sit for 15-20 minutes before running a short cycle using the hottest water. The baking soda will further absorb the unpleasant odors and will leave your dishwasher smell fresh.

7 0

2 years ago

A strings. A 50 gram bullet, traveling horzontally, strikes the block and becomes embedded inside the block. Immediately after t

vazorg [7]

Answer:

u= 200 m/s

Explanation:

Given that

Mass of bullet ,m= 50 gm

Assume that mass of block ,M= 1.2 kg

Lets take speed of the bullet before collision = u m/s

The speed of the system after collision ,v= 8 m/s

There is no any external force ,that is why linear momentum of the system will be conserve.

Linear momentum ,P = mass x velocity

m u = (M+m)v

0.05 x u = (1.2 + 0.05 ) x 8

u= 200 m/s

Therefore the speed of the bullet just before the collision is 200 m/s.

5 0

3 years ago

A bullet with mass m = 0.1 kg grams hits a ballistic pendulum with length L = 3 meters and mass M = 2 kg and lodges in it. When

Mnenie [13.5K]

Answer:

e) v₁ = 29.7 m / s

Explanation:

Let's propose the solution of the problem, let's start at the moment

Initial

p₀ = m v₁

Final

$p_{f}$ = (m + M) v

The moment is preserved

p₀ = $p_{f}$

m v₁ = (m + M) v

v = m / (m + M) v₁ (1)

a) energy is conserved

Let's look for kinetic energy

Initial

K₀ = ½ m v₁²

Final

$K_{f}$ = ½ (m + M) v²

Let's replace v

$K_{f}$ = ½ (m + M) [m / (m + M) v₁]²

$K_{f}$ = ½ m² / (m + M) v₁²

Let's look for the relationship of these energies

Ko / $K_{f}$ = ½ m v₁² / (½ m² / (m + M) v₁²)

Ko / $K_{f}$ = (m + M) / m = (1 + M / m)²

The kinetic energy changes therefore it is not conserved in the process, the missing energy is converted into potential heat energy during the crash

b) The impulse is conserved because the system is defined as formed by the two bodies and the externals are of action and reaction, so for the complete system the sum is zero and the moment does not change in value

c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential

d) when the pendulum oscillates the speed changes from v to zero, so the moment is not conserved, this is because there is an external force acting on the system, the force of gravity

e) For this part let's start at the end of the movement

It is system (bullet + block) moves, energy is conserved

Final. Highest point

$Em_{f}$ = U = (m + M) g h

Initial. Lowest point

Em₀ = K = ½ (m + M) v2

Em₀ = $Em_{f}$

½ (m + M) v² = (m + M) g h

v = √ 2gh

Let's look for the height (h) by trigonometry

Cos 15 = x / L

h = L-x

h = L - L cos 15

h = L (1- cos 15)

We replace

v = √ (2gL (1- cos 15))

Now we use equation (1) of momentum conservation

v = m / (m + M) v1

v₁ = (m + M) / m v

v1 = (0.1 +2.0) /0.1 RA (2 9.8 3 (1- cos 15))

v₁ = 21 √ (2.00)

v₁ = 29.7 m / s

6 0

3 years ago