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3 years ago

Which group contains only nonmetals

Physics

2 answers:

jolli1 [7]3 years ago

8 0

Noble gases, halogens, and non-metals (that is literally the group name)

vaieri [72.5K]3 years ago

6 0

<span>Group 14 is the first which contains non-metals. The non-metal it contains is carbon. </span>

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During a storm, the power went out at Bailey’s house. She had no electricity. What could she use as an alternate form of electri

evablogger [386]

C a battery-operated lamp sorry if incorrect

4 0

4 years ago

Read 2 more answers

In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 2.15 kg. The

Maksim231197 [3]

Answer:

0.339 kgm²

Explanation:

We know the period of this pendulum, T = 2π√(I/mgh) where I = moment of inertia of the object about the pivot axis, m = mass of object = 2.15 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.163 m.

Since T = 2π√(I/mgh), making I subject of the formula, we have

I = mghT²/4π²

Now since it takes 241 s to complete 113 cycles, then it takes 241 s/113 cycles to complete one cycle.

So, T = 241 s/113 = 2.133 s

So, Substituting the values of the variables into I, we have

I = mghT²/4π²

I = 2.15 kg × 9.8 m/s² × 0.163 m × (2.133 s)²/4π²

I = 15.63/4π² kgm²

I = 0.396 kgm²

Now from the parallel axis theorem, I = I' + mh² where I' = moment of inertia of object with respect to its center of mass about an axis parallel to the pivot axis

I' = I - mh²

I' = 0.396 kgm² - 2.15 kg × (0.163 m)²

I' = 0.396 kgm² - 0.057 kgm²

I' = 0.339 kgm²

7 0

3 years ago

Problem 7.17 A horizontal spring with stiffness 0.8 N/m has a relaxed length of 12 cm (0.12 m). A mass of 23 grams (0.023 kg) is

Jlenok [28]

The speed of the mass v = 0.884 m/s.

<u>Explanation</u>:

Let

K1 represents the kinetic energy of the mass when it is released,

U1 represents the potential energy of the spring when the mass is released,

K2 represents the kinetic energy of the mass when the spring returns to relaxed length,

U2 represents the potential energy of the spring when the spring returns to relaxed length

The spring is stretched by 0.27 - 0.12 = 0.15 m

K1 = 0

U1 = (1/2) $\times$ 0.8 $\times$ (0.15)^2

= 0.009 J

U2 = 0

By conservation of energy,

K2 + U2 = K1 + U1

K2 + 0 = 0 + 0.009 J

K2 = 0.009 J

Let v = speed of the mass

K2 = 1/2 $\times$ m $\times$ v^2

m = 23 g = 0.023 kg

0.009 = 1/2 $\times$ 0.023 $\times$ v^2

0.009 = 0.0115 $\times$ v^2

v = √(0.009 / 0.0115)

v = 0.884 m/s.

5 0

4 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

bogdanovich [222]

Answer:

Toyota Camry

F_d = 51.852 N

F_d = 100.042 N

Hummer H2

F_d = 412.0888 N

F_d = 8351.755 N

Explanation:

Given:

- The density of air p_air = 1.2 kg/m^3

- The drag force equals car's engine force.

Find:

- What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.)

- What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.)

Solution:

- The formula for drag force is given as follows:

F_d = 0.5*C_d*p_air*A*V^2

Where,

A : The drag Area m^2

C_d: The drag coefficient

V: Velocity m/s

a) Toyota Camry

C_d = 0.28 , A = 0.70 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

F_d = 0.5*0.28*1.2*0.70*(22.22)^2

F_d = 51.852 N

C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

F_d = 0.5*0.28*1.2*0.70*(29.1667)^2

F_d = 100.042 N

b) Hummer H2

C_d = 0.57 , A = 2.44 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

F_d = 0.5*0.57*1.2*2.44*(22.22)^2

F_d = 412.0888 N

C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

F_d = 0.5*0.57*1.2*2.44*(29.1667)^2

F_d = 8351.755 N

4 0

3 years ago

You take a bucket outside and hold it as you normally would to hold water. As the rain comes down, the flux of water out of the

Brums [2.3K]

Answer:

the bucket-surface area vector and the direction of the rain is 180° which gives a negative flux in the outward direction.

If we cut the base of the bucket then hold it in the rain then the there is no area intercepted by the rainfall and hence we have a zero flux.

Explanation:

We know that flux is defined as the density of any energy field passing through a given area parallel to the area vector.

Mathematically:

$\phi=\bar E.A\cos\theta$

where:

$\phi=$ flux through the area

$A=$ area of concern

$\bar E=$ density of energy field per unit area

$\theta=$ angle between the energy field lines and the normal to area.

When we hold a bucket in the rain then assuming that the base of the bucket is normal to the direction of rainfall then the angle between the area vector and the rain drop is zero when observed from inside.

For the flux outside the bucket we are concerned with the outer surface of the bucket, now the angle formed between the bucket-surface area vector and the direction of the rain is 180° which gives a negative flux in the outward direction.

If we cut the base of the bucket then hold it in the rain then the there is no area intercepted by the rainfall and hence we have a zero flux.

6 0

3 years ago