What is tan θ for the given triangle?

An object at rest cannot remain at rest unless which of the following holds?View Available Hint(s)An object at rest cannot remai

Leni [432]

Answer:

True The net force must be zero for the acceleration to be zero

Explanation:

In order to analyze the statements of this problem we propose your solution.

First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)

Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.

As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.

Now we can analyze the statements given.

True The net force must be zero for the acceleration to be zero

False. If the force is different from zero, there is acceleration that changes the speeds

False. There may be forces, but the sum of them must be zero

False. If a force acts, the acceleration is different from zero and the speed changes

5 0

3 years ago

Vector vector b has x, y, and z components of 4.00, 4.00, and 2.00 units, respectively. calculate the magnitude of vector

Sav [38]

Good morning.

We see that $\mathsf{\overset{\to}{b}} = \mathsf{(4.00, \ 4.00, \ 2.00)}$

The magnitude(norm, to be precise) can be calculated the following way:

$\star \ \boxed{\mathsf{\overset{\to}{a}=(x, y,z)\Rightarrow ||\overset{\to}{a}|| = \sqrt{x^2+y^2+z^2}}}$

Now the calculus is trivial:

$\mathsf{\|\overset{\to}{b} \| =\sqrt{4^2+4^2+2^2} =\sqrt{16+16+4}}\\ \\ \mathsf{\|\overset{\to}{b}\|=\sqrt{36}}\\ \\ \boxed{\mathsf{\|\overset{\to}{b}\| = 6.00 \ u}}$

7 0

3 years ago

A truck is moving around a circular curve at a uniform velocity of 13 m/s. If the centripetal force on the truck is 3,300 N and

Paladinen [302]

Answer: Option B.

Since here the truck is moving on a circular track, it will experience centripetal force.

F(centripetal) = m × acc
or
$r = \frac{m v^{2}}{F}$

where r is the radius of the track.
m is the mass of truck
v is the speed of the truck.
Given: v = <span>13 m/s
m = </span><span>1,600 kg
</span>F = 3300 Newton

To find = radius of track=?
$r = \frac{m v^{2} }{F}$
$r = \frac{1600*13*13}{3300}$
r = 81.94 m
Therefore, radius of track is 81.94 m

4 0

3 years ago

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Difference between upthrust force and normal reaction force.

irakobra [83]

Answer:

Explanation:

Both are contact forces arising at the interface between two bodies. In the fluid this interface might be irregular, and it completely surrounds a submerged object. For a solid it is usually a single flat surface - but it can be a collection of surfaces, which do not need to be flat or regular, and which can surround the object

Upthrust occurs at a fluid-solid interface whereas normal reaction occurs at a solid-solid surface. However, it is possible to generate the same fluid-like phenomenon of upthrust by immersing a solid object in sand or small beads and agitating them to simulate the pressure of atoms. With

7 0

2 years ago

A simple harmonic oscillator consists of a block (m = 0.50 kg) attached to a spring (k = 128 N/m). The block is pulled a certain

Zinaida [17]

Answer:

0.5 m

14.00595

8 m/s, 0.0625 s

5.71314 m/s

Explanation:

k = Spring constant = 128 N/m

A = Amplitude

E = Energy in spring = 16 J

Energy in spring is given by

$E=\dfrac{1}{2}kA^2\\\Rightarrow A=\sqrt{\dfrac{2E}{k}}\\\Rightarrow A=\sqrt{\dfrac{2\times 16}{128}}\\\Rightarrow A=0.5\ m$

The amplitude is 0.5 m

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{0.5}{128}}\\\Rightarrow T=0.39269\ s$

Number of oscillations is given by

$N=\dfrac{5.5}{0.39269}\\\Rightarrow N=14.00595$

The number of oscillations is 14.00595

For maximum speed

$\dfrac{1}{2}mv^2=16\\\Rightarrow v=\sqrt{\dfrac{16\times 2}{0.5}}\\\Rightarrow v=8\ m/s$

The maximum speed is 8 m/s

For a distance of 0.5 m which is the amplitude

$Time=\dfrac{Distance}{Speed}\\\Rightarrow Time=\dfrac{0.5}{8}\\\Rightarrow Time=0.0625\ s$

The time taken would be 0.0625 s

The maximum kinetic energy is equal to the mechanical energy

$\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=16$

At x = 0.35 m

$v=\sqrt{\dfrac{16-\dfrac{1}{2}kx^2}{\dfrac{1}{2}m}}\\\Rightarrow v=\sqrt{\dfrac{16-\dfrac{1}{2}128\times 0.35^2}{\dfrac{1}{2}0.5}}\\\Rightarrow v=5.71314\ m/s$

The speed of the block is 5.71314 m/s

4 0

3 years ago

Read 2 more answers