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3 years ago

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A scientist is studying a sample of matter the matter has no definite shape or volume and it can conduct an electric current. Th

e sample is most likely in which of the following states
A.solid
B.Liquid
C.gas
D.plasma

2 answers:

boyakko [2]3 years ago

7 0

it will be d hope i help

Anna35 [415]3 years ago

4 0

D. plasma ..................

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The four-wheel-drive all-terrain vehicle has a mass of 385 kg with center of mass G2. The driver has a mass of 75 kg with center

jenyasd209 [6]

The coefficient of friction is missing and it has a value of μ = 0.4

Answer:

a = 3.924 m/s²

Explanation:

I've attached the kinematic free body diagram.

Taking the sum of all upward and downward forces,

EFy = 0;

N1 + N2 - m_p•g - m_v•g = 0

N1 + N2 = m_p•g + m_v•g

Where;

N1 and N2 are the normal reactions at the wheels

m_p is the mass of the driver

m_v is the mass of the vehicle

g is the acceleration due to gravity.

Plugging in the relevant values in the question,we obtain;

N1 + N2 = (385 + 75) x 9.81

N1 + N2 = 4512.6N - - - (eq1)

Now, taking sum of all horizontal forces;

EFx = (m_p + m_v) x a

So,

μ(N1 + N2) = (mp + mv) x a

Thus,

0.4(N1 + N2) = (385 + 75)a

0.4(N1 + N2) = 460a

N1 + N2 = 1150a

From eq(1),N1 + N2 = 4512.6N

Thus,

1150a = 4512.6N

a = 4512.6/1150

a = 3.924 m/s²

Therefore, the acceleration, a = 3.924 m/s²

7 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo

Alla [95]

Answer:

$v_{ox}= 19.6\ m/s$

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = $v_{ox}$

Initial vertical velocity= $v_{oy}$ (Since ball was thrown horizontally only)

Acceleration acting horizontally, $a_x$ = 0 m/s² [ Since no acceleration acts horizontally) ]

Vertical Acceleration, $a_y$ = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

$H= v_{oy}t+\frac{1}{2}a_yt^2$

$5= (0)t+\frac{1}{2}(9.8)t^2$

$t= \frac{10}{9.8}=1.02 s$

Then using Equations of motion for horizontal motion,

$R= v_{ox}t+\frac{1}{2}a_xt^2$

$20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2$

$v_{ox}= 19.6\ m/s$

4 0

3 years ago

How long will it take the officer to catch the thief?

jenyasd209 [6]

Answer:

can take a lot of time

Time is money

Explanation:

3 0

4 years ago

A satellite is in a circular orbit around Mars, which has a mass M = 6.40 × 1023 kg and radius R = 3.40 ×106 m.

Pepsi [2]

Answer:

a) The orbital speed of a satellite with a orbital radius R (in meters) will have an orbital speed of approximately $\displaystyle \sqrt\frac{4.27 \times 10^{13}}{R}\; \rm m \cdot s^{-1}$ .

b) Again, if the orbital radius R is in meters, the orbital period of the satellite would be approximately $\displaystyle 9.62 \times 10^{-7}\, R^{3/2}\; \rm s$ .

c) The orbital radius required would be approximately $\rm 2.04 \times 10^7\; m$ .

d) The escape velocity from the surface of that planet would be approximately $\rm 5.01\times 10^3\; m \cdot s^{-1}$ .

Explanation:

<h3>a)</h3>

Since the orbit of this satellite is circular, it is undergoing a centripetal motion. The planet's gravitational attraction on the satellite would supply this centripetal force.

The magnitude of gravity between two point or spherical mass is equal to:

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}}$ ,

where

$G$ is the constant of universal gravitation.
$M$ is the mass of the first mass. (In this case, let $M$ be the mass of the planet.)
$m$ is the mass of the second mass. (In this case, let $m$ be the mass of the satellite.)
$r$ is the distance between the center of mass of these two objects.

On the other hand, the net force on an object in a centripetal motion should be:

$\displaystyle \frac{m \cdot v^{2}}{r}$ ,

where

$m$ is the mass of the object (in this case, that's the mass of the satellite.)
$v$ is the orbital speed of the satellite.
$r$ is the radius of the circular orbit.

Assume that gravitational force is the only force on the satellite. The net force should be equal to the planet's gravitational attraction on the satellite. Equate the two expressions and solve for $v$ :

$\displaystyle \frac{G \cdot M \cdot m}{r^{2}} = \frac{m \cdot v^{2}}{r}$ .

$\displaystyle v^2 = \frac{G \cdot M}{r}$ .

$\displaystyle v = \sqrt{\frac{G \cdot M}{r}}$ .

Take $G \approx 6.67 \times \rm 10^{-11} \; m^3 \cdot kg^{-1} \cdot s^{-2}$ , Simplify the expression $v$ :

$\begin{aligned} v &= \sqrt{\frac{G \cdot M}{r}} \cr &= \sqrt{\frac{6.67 \times \rm 10^{-11} \times 6.40 \times 10^{23}}{r}} \cr &\approx \sqrt{\frac{4.27 \times 10^{13}}{r}} \; \rm m \cdot s^{-1} \end{aligned}$ .

<h3>b)</h3>

Since the orbit is a circle of radius $R$ , the distance traveled in one period would be equal to the circumference of that circle, $2 \pi R$ .

Divide distance with speed to find the time required.

$\begin{aligned} t &= \frac{s}{v} \cr &= 2 \pi R}\left/\sqrt{\frac{G \cdot M}{R}} \; \rm m \cdot s^{-1}\right. \cr &= \frac{2\pi R^{3/2}}{\sqrt{G \cdot M}} \cr &\approx 9.62 \times 10^{-7}\, R^{3/2}\; \rm s\end{aligned}$ .

<h3>c)</h3>

Convert $24.6\; \rm \text{hours}$ to seconds:

$24.6 \times 3600 = 88560\; \rm s$

Solve the equation for $R$ :

$9.62 \times 10^{-7}\, R^{3/2}= 88560$ .

$R \approx 2.04 \times 10^7\; \rm m$ .

<h3>d)</h3>

If an object is at its escape speed, its kinetic energy (KE) plus its gravitational potential energy (GPE) should be equal to zero.

$\displaystyle \text{GPE} = -\frac{G \cdot M \cdot m}{r}$ (Note the minus sign in front of the fraction. GPE should always be negative or zero.)

$\displaystyle \text{KE} = \frac{1}{2} \, m \cdot v^{2}$ .

Solve for $v$ . The value of $m$ shouldn't matter, for it would be eliminated from both sides of the equation.

$\displaystyle -\frac{G \cdot M \cdot m}{r} + \frac{1}{2} \, m \cdot v^{2}= 0$ .

$\displaystyle v = \sqrt{\frac{2\, G \cdot M}{R}} \approx 5.01\times 10^{3}\; \rm m\cdot s^{-1}$ .

5 0

4 years ago

Name two laboratory equipment in the picture . b ) Identify two mistakes that these people in the laboratory above are doing

pishuonlain [190]

The two laboratory equipment in the picture are beaker and chemicals.The people in the picture are not wearing any protective gears while (a) working in the laboratory (b) Throwing the chemicals on the floor.

Among the many items that would be considered general lab equipment are pipettes, scales, centrifuges, Bunsen burners, freezers, hot plates, incubators, coolers, stirrers, water baths, and fume hoods

The dangers of working in a laboratory setting include:

Chemical hazards: Handling toxic substances can cause irritation and carcinogenicity.

Biological hazards: Biological hazards include hazards from working with small animals, working with blood borne pathogens and working with biological agents, such as viruses and bacteria.

Physical hazards: Physical hazards include exposure to noise, poor posture and the explosibility and flammability of substances.

Safety hazards: Safety hazards include unbalanced centrifuges, danger when handling hot sterilized items and electrical hazards, such as shock, explosions, blasts and electrocutions.

Allergy hazards: A common allergy hazard in the laboratory setting is a latex allergy, as many of the materials used in a laboratory setting are latex.

Learn more about Laboratory experiments at brainly.com/question/8430576

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1 year ago