A bus travelling 30.0 km/h east has a constant increase in speed of 1.5 m/s. What is its velocity 6.8s later

MGS and later orbiters found spectral evidence for what minerals on the Martian surface? Check all that apply.

slamgirl [31]

Mars Global Surveyors (MGS) and later orbiters found the following minerals on the Martian surface;

Phyllosilicates

Carbonate
Sulfates
Iron oxide

The Mars Global Surveyors (MGS) and later orbiters suggest that the Martian crust contains a higher percentage of volatile elements such as Sulphur and chlorine than the Earth's crust does.

These scientists also conclude that the most abundant chemical elements in the Martian crust are those found in Igneous rock.

These elements include the following;

Silicon,
Oxygen,
Iron,
Magnesium,
Aluminum,
Calcium, and
Potassium.

They also, suggest that hydrogen is found in ice (water) while carbon is found in carbon dioxide and carbonates.

From the given options the minerals found in Martian surface include;

Phyllosilicates ------ these are sheet of silicate minerals

Carbonate
Sulfates
iron oxide

Learn more here: brainly.com/question/20470323

6 0

2 years ago

If a 5.5 kg object experiences 15 N of force for .15 seconds what is the speed change

Nady [450]

The speed change : Δv = 0.41 m/s

<h3>Further explanation</h3>

Given

mass = 5.5 kg

Force = 15 N

time = 0.15 s

Required

the speed change

Solution

Newton 2nd's law

Impulse and momentum

F = m.a

F = m . Δv/t

F.t = m.Δv

Input the value :

15 N x 0.15 s = 5.5 kg x Δv

Δv = 0.41 m/s

8 0

3 years ago

Witch of the following is not an appropriate unit for power

erica [24]

There are no appropriate units for power on the list you provided

6 0

3 years ago

A bug is 12 cm from the center of a turntable that is rotating with a frequency of 45 rev/min . What minimum coefficient frictio

Agata [3.3K]

Answer:

The minimum coefficient of friction is 0.27.

Explanation:

To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.

Centripetal force is written as

$F_c = m\frac{v^2}{r}$

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

$v = r\omega$

with ω denoting the angular velocity, which we are given. With that, the above becomes:

$F_c = m\frac{v^2}{r}=m\omega^2 r$

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

$F_r=\mu mg \geq m\omega^2 r = F_c\implies\\\mu \geq \frac{\omega^2 r}{g}$

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

$\frac{1 rev}{1 min}\cdot\frac{\frac{2\pi rad}{rev}}{\frac{60s}{1 min}}=\frac{2\pi}{60}\frac{rad}{s}$

and so 45 rev/min = 4.71 rad/s.

$\mu \geq \frac{\omega^2 r}{g}=\frac{4.71^2\frac{1}{s^2}\cdot 0.12m}{9.8\frac{m}{s^2}}=0.27$

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.

3 0

3 years ago

Raj is trying to make a diagram to show what he has learned about nuclear fusion.

KIM [24]

No, he should place the He atom and energy on the right, and the H atoms and the heat and energy on the left.

6 0

3 years ago

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