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scoray [572]
4 years ago
13

I need physics help!!

Physics
1 answer:
nlexa [21]4 years ago
6 0

Answer: magnitude = 5 and angle = 160°

Explanation:

A vector has magnitude 5 and angle -20°

A vector can be written in terms of its components as :

\vec{A}=A_x\hat{i}+A_y\hat j \\ \Rightarrow \vec A = A cos\theta \hat i+A sin\theta \hat j

using trigonometry identity:

sin θ = -sin ( 180 + θ )

cos θ = - cos (180 + θ)

Thus, we can write the same vector as:

\vec A = -A cos(180+\theta) \hat i+(-A) sin(180+\theta) \hat j

Magnitude |A|=\sqrt{(-A)^2+(-A)^2} = A = 5

Thus, the magnitude would be A = 5 and angle = 180+θ= 180-20 = 160°

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Which type of reaction does this diagram represent?
Ugo [173]

The diagram represents a chain reaction that is caused by nuclear fission.

<h3>What is a nuclear fission reaction?</h3>

A nuclear fission reaction is a reaction in which the nucleus of a larger atom is split into two or more smaller nucleus of atoms.

Nuclear fission can proceed in the form of a chain reaction in which the products of the first fission reaction are used to initiate further fission reactions.

Therefore, the diagram represents a chain reaction that is caused by nuclear fission.

Learn more about nuclear fission at: brainly.com/question/22155336

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4 0
2 years ago
Read 2 more answers
On what factors, current sensitivity voltage sensitivity of<br>a galvanometer depend?​
Eddi Din [679]

Answer:

Current sensitivity voltage sensitivity of a galvanometer depend is on a number of turns, coil area, restoring force per unit twist and megnetic filed.

7 0
3 years ago
While walking between gates at an airport, you notice a child running along a moving walkway. Estimating that the child runs at
Mashcka [7]

Answer:

The speed of the moving walkway is 1.50 m/s

Explanation:

The position of the child can be calculated using the following equation:

x = x0 + v · t

Where :

x = position of the child at time t.

v = velocity of the child.

t = time.

When the child runs in the same direction as the walkway, the velocity of the child will be its  velocity relative to the walkway plus the velocity of the walkway. Then, if we place the origin of the frame of reference at the start of the walkway:

x = x0 + v · t

25 m = 0 m + (2.8 m/s + v) · t₁

Where v is the velocity of the walkway

On its way back, the velocity of the child relative to the walkway is in the opposite direction to the velocity of the walkway. Then:

x = x0 + v · t

0 m = 25 m + (-2.8 m/s + v) · t₂

We also know that t₁ + t₂ = 25 s

Then: t₁ = 25 - t₂

So, we can write the following system of equations:

25 m = (2.8 m/s + v) · (25 s - t₂)

-25 m = (-2.8 m/s + v) · t₂

Let´s take the second equation and solve it for t₂

-25 m / (-2.8 m/s + v) = t₂

Now, let´s replace t₂ in the first equation:

25 m = (2.8 m/s + v) · (25 s + 25 m / (-2.8 m/s + v))

Let´s sum the fraction: 25 s + 25 m / (-2.8 m/s + v)

25 m = (2.8 m/s + v) · (25 s ·(-2.8 m/s + v) + 25 m) / (-2.8 m/s + v)

multiply by (-2.8 m/s + v) both sides of the equation:

25 m(-2.8 m/s + v) = (2.8 m/s + v) · (-70 m + 25 s · v + 25 m)

Apply distributive property:

-70 m²/s +25 m·v = -196 m²/s +70 m·v +70 m²/s -70 m·v +25 s ·v² + 25 m v

56 m²/s = 25 s · v²

56 m²/s / 25 s = v²

v = 1.50 m/s

The speed of the moving walkway is 1.50 m/s

7 0
3 years ago
A 0.300 kg block is pressed against a spring with a spring constant of 8050 N/m until the spring is compressed by 6.00 cm. When
natita [175]

Answer:

a) \mu_{k} = 0.704, b) R = 0.312\,m

Explanation:

a) The minimum coeffcient of friction is computed by the following expression derived from the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \mu_{k}\cdot m\cdot g \cdot \Delta s

\mu_{k} = \frac{k\cdot x^{2}}{2\cdot m\cdot g \cdot \Delta s}

\mu_{k} = \frac{\left(8050\,\frac{N}{m} \right)\cdot (0.06\,m)^{2}}{2\cdot (0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (7\,m)}

\mu_{k} = 0.704

b) The speed of the block is determined by using the Principle of Energy Conservation:

\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}

v = x\cdot \sqrt{\frac{k}{m} }

v = (0.06\,m)\cdot \sqrt{\frac{8050\,\frac{N}{m} }{0.3\,kg} }

v \approx 9.829\,\frac{m}{s}

The radius of the circular loop is:

\Sigma F_{r} = -90\,N -(0.3\,kg)\cdot (9.807\,\frac{m}{s^{2}} ) = -(0.3\,kg)\cdot \frac{v^{2}}{R}

\frac{\left(9.829\,\frac{m}{s}\right)^{2}}{R} = 309.807\,\frac{m}{s^{2}}

R = 0.312\,m

5 0
4 years ago
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lys-0071 [83]

Answer:

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7 0
2 years ago
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