A car's speed 3 seconds after accelerating from rest at 2 m/s^2 is?

Explain how we use a spectrometer to help us determine what stars and planets are made of?

BaLLatris [955]

We use a spectrometer to help us determine what stars and planets are made of by passing a light through various chemical elements, different spectral patterns are created. By matching those patterns up to patterns generated in a laboratory environment we can tell what the composition of a distant star or planet is.

4 0

4 years ago

A 5.50 kg sled is initially at rest on a frictionless horizontal road. The sled is pulled a distance of 3.20 m by a force of 25.

kiruha [24]

(a) 69.3 J

The work done by the applied force is given by:

$W=Fd cos \theta$

where:

F = 25.0 N is the magnitude of the applied force

d = 3.20 m is the displacement of the sled

$\theta=30^{\circ}$ is the angle between the direction of the force and the displacement of the sled

Substituting numbers into the formula, we find

$W=(25.0 N)(3.20 m)(cos 30^{\circ})=69.3 J$

(b) 0

The problem says that the surface is frictionless: this means that no friction is acting on the sled, therefore the energy dissipated by friction must be zero.

(c) 69.3 J

According to the work-energy theorem, the work done by the applied force is equal to the change in kinetic energy of the sled:

$\Delta K = W$

where

$\Delta K$ is the change in kinetic energy

W is the work done

Since we already calculated W in part (a):

W = 69.3 J

We therefore know that the change in kinetic energy of the sled is equal to this value:

$\Delta K=69.3 J$

(d) 4.9 m/s

The change in kinetic energy of the sled can be rewritten as:

$\Delta K=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$ (1)

where

Kf is the final kinetic energy

Ki is the initial kinetic energy

m = 5.50 kg is the mass of the sled

u = 0 is the initial speed of the sled

v = ? is the final speed of the sled

We can calculate the variation of kinetic energy of the sled, $\Delta K$ , after it has travelled for d=3 m. Using the work-energy theorem again, we find

$\Delta K= W = Fd cos \theta =(25.0 N)(3.0 m)(cos 30^{\circ})=65.0 J$

And substituting into (1) and re-arrangin the equation, we find

$v=\sqrt{\frac{2 \Delta K}{m}}=\sqrt{\frac{2(65.0 J)}{5.50 kg}}=4.9 m/s$

6 0

4 years ago

8.

Leno4ka [110]

Gravitational force is an example of at-a-distance force

Option: C

Explanation:

Force has both direction and magnitude it is a "vector" quantity. Force is defined as pull or push of an object resulting it to interact with two objects. There are 'Contact forces' and 'At-a-distance forces'. If two interacting objects are not in a physical contact with respect to each other the force is exerted between these two objects called 'at-a-distance' forces. This type of forces consists gravitational forces. Some of the examples of the 'at-a-distance' forces are electrical and magnetic.

8 0

4 years ago

In projectile motion, why don't we add acceleration due to gravity on the formula of horizontal motion

Basile [38]

Answer:

Horizontal motion is free from gravity which is zero

8 0

3 years ago

Read 2 more answers

What is the mass of a 10.4 cm³ cube of lead.

Gekata [30.6K]

<h2>Answer:</h2>

The mass of a 10.4 cm³ is, approximately, 117.42 grams.

<h2>Explanation:</h2>

Volume of the cube: 10.4 cm³

Now, let's look up the density of lead:

Density of lead: 11.29 g/cm³.

This value fo density tells us that 1 cm³ of lead has a mass of 11.29g. Hence, we can stablish the following rule of 3 to find the mass of our cube.

1 cm³ ----------------> 11.29 g

10.4 cm³------------> x

x= (10.4 cm³ ⋅ 11.29 g)/1 cm³

x= 117.42 g.

Conclusion.

The mass of a 10.4 cm³ is, approximately, 117.42 g.

7 0

2 years ago