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Inga [223]
3 years ago
15

What is not an example of a mineral?a.gold b.mica c.steel d.quartz

Physics
2 answers:
vodka [1.7K]3 years ago
7 0

Answer:

C  steel

Explanation:

Leviafan [203]3 years ago
6 0

Answer:

C : Steel

Explanation:

Steel is manufactured and not made by the Earth itself

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A boy throws a baseball onto a roof and it rolls back down and off the roof with a speed of 3.75 m/s. If the roof is pitched at
dalvyx [7]

Answer:

a) t = 0.528 s

b) D = 1.62 m

Explanation:

given,

speed of the baseball = 3.75 m/s

angle made with the horizontal = 35°

height of the roof edge = 2.5 m

using equation of motion

s = ut +\dfrac{1}{2}gt^2

2.5 = vsin\theta \ t +\dfrac{1}{2}gt^2

2.5 = 3.75\ sin35^0 \ t +\dfrac{1}{2}\times 9.8 t^2

4.9 t² + 2.15 t - 2.5 = 0

on solving the above equation

t = 0.528 s

b) D = v cos θ × t

D = 3.75 × cos 35° ×0.528

D = 1.62 m

3 0
4 years ago
When you see a fish in the water where is it really ?
photoshop1234 [79]

It really depends on the angle where you look at it from and what type of glass/shape they are in. Mine always appeared pretty close even when it wasn't.


Source: Had 5 fish of my own.


Have a lovely day! ~Pooch ♥

6 0
3 years ago
A exercise ball is on a table 2.5 m above the ground. What is the mass
horsena [70]

Answer:

2.27kg

Explanation:

The Potential energy gained = Force gravity× Height

Force gravity on the ball is GPE/ height

56.8/2.5=22.72N

the mass = 22.72/10= 2.27kg

4 0
3 years ago
An insect meanders across a sidewalk. The insect moves 15cm to the right, 10cm up the sidewalk and 8cm to the left. What is the
IRISSAK [1]

Answer:12.206 cm,\theta =54.99^{\circ}

Explanation:

Given

Insect walks 15 cm to the right

so its position vector isr_1=15i

Now it moves 10 cm up so its new position vector

r_2=15i+10j

Now it moves 8 cm left so its final position vector is

r_3=15\hat{i}+10\hat{j}-8\hat{i}=7\hat{i}+10\hat{j}

so its displacement is given by

|r_3|=\sqrt{7^2+10^2}=\sqrt{149}=12.206 cm

For direction, let \theta is the angle made by its position vector with x axis

tan\theta =\frac{10}{7}=1.428

\theta =54.99^{\circ}

8 0
3 years ago
Suppose a car approaches a hill and has an initial speed of 108 km/h at the bottom of the hill. The driver takes her foot off of
Aleks04 [339]

Answer:

a) The car will reach a height of 45.9 m.

b) The amount of thermal energy generated is 173382 J.

c) The magnitude of the force of friction is 417.8 N.  

Explanation:

Hi there!

a) In this problem, we have to use the conservation of energy. The energy conservation theorem states that the energy of a system remains constant. Energy can´t be created nor destroyed, only transformed. In the case of the car, the initial kinetic energy is transformed into potential energy as the car´s height increases while coasting up the hill.

Then, all the initial kinetic energy (KE) will be transformed into potential energy (PE) (only if there is no friction).

The equation of KE is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

The equation of PE is the following:

PE = m · g · h

Where:

m = mass of the car.

g = acceleration due to gravity.

h = height at which the car is located.

Since work done by friction is negligible, we can assume that all the initial kinetic energy will be transformed into potential energy. Then:

KE at the bottom of the hill = PE at the top of the hill

1/2 · m · v² = m · g · h

Solving for h:

1/2 · v² / g = h

Let´s convert the speed unit into m/s:

108 km/h · 1000 m/ 1 km · 1 h / 3600 s = 30 m/s

Now, let´s calculate h:

h = 1/2 · (30 m/s)² / 9.8 m/s²

h = 45.9 m

The car will reach a height of 45.9 m.

b) In this case, all the kinetic energy is not transformed into potential energy because some energy is transformed into thermal energy due to friction. The thermal energy generated is equal to the work done by friction. Then:

KE at the bottom of the hill = PE + work done by friction

KE = PE + Wfr  (where Wfr is the work done by friction).

1/2 · m · v² = m · g · h + Wfr

1/2 · m · v² - m · g · h = Wfr

1/2 · 710 kg · (30 m/s)² - 710 kg · 9.8 m/s² · 21 m = Wfr

Wfr = 173382 J

The amount of thermal energy generated is 173382 J.

c) The work done by friction is calculated as follows:

Wfr = Ffr · Δx

Where:

Ffr = friction force.

Δx = traveled distance

Please, see the attached figure to notice that the traveled distance can be calculated by trigonometry using this trigonometric rule of right triangles:

sin angle = opposite side / hypotenuse

In our case:

sin 2.9° = h / Δx

Δx = h / sin 2.9°

Δx = 21 m / sin 2.9° = 415 m

Then, solving for the friction force using the equation of the work done by friction:

Wfr = Ffr · Δx

Wfr / Δx = Ffr

173382 J / 415 m = Ffr

Ffr = 417.8 N

The magnitude of the force of friction is 417.8 N

6 0
3 years ago
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