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2 years ago

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a scooter has a mass of 25 kg. a constant force is exerted on it for 5 s during the time the force is exerted the scooter increa

ses its speed from 6 m/s to 26 m/s what is the magnitude of the force exerted on the scooter

1 answer:

Afina-wow [57]2 years ago

5 0

Answer:

Answer was deleted first time, the answer is 917 N

Explanation:

Hope this helps!

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A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

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

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

3 years ago

In a game of pool, the cue ball moves at a speed of 2 m/s toward the eight ball. When the cue ball hits the eight ball, the cue

agasfer [191]

Answer:

a) p₀ = 1.2 kg m / s, b) p_f = 1.2 kg m / s, c) θ = 12.36, d) v_{2f} = 1.278 m/s

Explanation:

a system formed by the two balls, which are isolated and the forces during the collision are internal, therefore the moment is conserved

a) the initial impulse is

p₀ = m v₁₀ + 0

p₀ = 0.6 2

p₀ = 1.2 kg m / s

b) as the system is isolated, the moment is conserved so

p_f = 1.2 kg m / s

we define a reference system where the x-axis coincides with the initial movement of the cue ball

we write the final moment for each axis

X axis

p₀ₓ = 1.2 kg m / s

p_{fx} = m v1f cos 20 + m v2f cos θ

p₀ = p_f

1.2 = 0.6 (-0.8) cos 20+ 0.6 v_{2f} cos θ

1.2482 = v_{2f} cos θ

Y axis

p_{oy} = 0

p_{fy} = m v_{1f} sin 20 + m v_{2f} cos θ

0 = 0.6 (-0.8) sin 20 + 0.6 v_{2f} sin θ

0.2736 = v_{2f} sin θ

we write our system of equations

0.2736 = v_{2f} sin θ

1.2482 = v_{2f} cos θ

divide to solve

0.219 = tan θ

θ = tan⁻¹ 0.21919

θ = 12.36

let's look for speed

0.2736 = v_{2f} sin θ

v_{2f} = 0.2736 / sin 12.36

v_{2f} = 1.278 m / s

7 0

2 years ago

As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over th

poizon [28]

Answer:

$4.048\times 10^{-7}\ m$

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

E = Energy = $4.91\times 10^{-19}\ J$

Wavelength ejected is given by

$\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m$

The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is $4.048\times 10^{-7}\ m$

4 0

3 years ago

What does Force Means

zubka84 [21]

Answer:

strength or energy as an attribute of physical action or movement.

Explanation:

coercion or compulsion, especially with the use or threat of violence.

this is from google

3 0

3 years ago

Read 2 more answers

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago