1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
harkovskaia [24]
2 years ago
15

Can anybody tell me the right answer ? please and thank you !!

Physics
1 answer:
ad-work [718]2 years ago
3 0

Answer:

c. selenium

Explanation:

sulfur and selenium are in the same group

You might be interested in
A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (v), measured in meters per second, by using this kinematic equation:

v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s} (1)

Where:

a - Acceleration, measured in meters per square second.

\Delta s - Travelled distance, measured in meters.

v_{o} - Initial velocity, measured in meters per second.

If we know that v_{o} = 0\,\frac{m}{s}, a = 2.35\,\frac{m}{s^{2}} and \Delta s = 595\,m, the final velocity of the rocket is:

v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}

v\approx 52.882\,\frac{m}{s}

The time associated with this launch (t), measured in seconds, is:

t = \frac{v-v_{o}}{a}

t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }

t = 22.503\,s

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o}) (2)

Where:

v_{o} - Initial speed, measured in meters per second.

v - Final speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

If we know that v_{o} = 52.882\,\frac{m}{s}, v = 0\,\frac{m}{s}, a = -9.807\,\frac{m}{s^{2}} and s_{o} = 595\,m, then the maximum height reached by the rocket is:

v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})

s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}

s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}

s = 737.577\,m

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2} (2)

Where:

s_{o} - Initial height, measured in meters.

s - Final height, measured in meters.

v_{o} - Initial speed, measured in meters per second.

a - Gravitational acceleration, measured in meters per square second.

t - Time, measured in seconds.

If we know that s_{o} = 595\,m, v_{o} = 52.882\,\frac{m}{s}, s = 0\,m and a = -9.807\,\frac{m}{s^{2}}, then the time needed by the rocket is:

0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}

-4.904\cdot t^{2}+52.882\cdot t +595 = 0

Then, we solve this polynomial by Quadratic Formula:

t_{1}\approx 17.655\,s, t_{2} \approx -6.872\,s

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0
2 years ago
Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST
Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

a^2+b^2=c^2

Plug values in

88.4^2+174^2=c^2

Simplify

7814.56+30276=c^2

Add the two values

38090.56=c^2

Take the square root of both sides

195.168\approx195.168

8 0
2 years ago
A sound wave has a speed of 345 m/s and a wavelength of 4.15 meters. What is the frequency of this wave?
Debora [2.8K]

Answer:

83 hertz

OPTION A

Explanation:

wavelength =  \frac{speed}{frequency}

Given

wavelength=4.15m

speed=345 m/s

frequency =  \frac{speed}{wavelength} =  \frac{345}{4.15}

frequency = 83 \: hertz

I hope it helped you

5 0
2 years ago
A 2.74 g coin, which has zero potential energy at the surface, is dropped into a 12.2 m well. After the coin comes to a stop in
VikaD [51]

Answer:

B. - 0.328

Explanation

Potential Energy:<em> This is the energy of a body due to position.</em>

<em>The S.I unit of potential energy is Joules (J).</em>

<em>It can be expressed mathematically as</em>

<em>Ep = mgh........................... Equation 1</em>

<em>Where Ep = potential energy, m = mass of the coin, h = height, g = acceleration due to gravity,</em>

<em>Given: m = 2.74 g = 0.00274 kg, h = 12.2 m, g = 9.8 m/s²</em>

Substituting these values into equation 1

Ep = 0.00274×12.2×9.8

Ep = 0.328 J.

Note: Since the potential energy at the surface is zero, the potential Energy with respect to the surface = -0.328 J

The right option is B. - 0.328

<em />

7 0
2 years ago
An 8 g bullet leaves the muzzle of a rifle with
Elena-2011 [213]

Answer: 1872 N

Explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

V^{2}=V_{o}^{2} + 2ad (1)

F=ma (2)

Where:

V=611.9 m/s is the bullet's final speed (when it leaves the muzzle)

V_{o}=0 is the bullet's initial speed (at rest)

a is the bullet's acceleration

d=0.8 m is the distance traveled by the bullet before leaving the muzzle

F is the force

m=8 g \frac{1 kg}{1000 g}=0.008 kg is the mass of the bullet

Knowing this, let's begin by isolating a from (1):

a=\frac{V^{2}}{2d} (3)

a=\frac{(611.9 m/s)^{2}}{2(0.8 m)} (4)

a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2} (5)

Substituting (5) in (2):

F=(0.008 kg)(2.34(10)^{5} m/s^{2}) (6)

Finally:

F=1872 N

4 0
2 years ago
Other questions:
  • A cylinder which is in a horizontal position contains an unknown noble gas at 4.63 × 104 Pa and is sealed with a massless piston
    11·1 answer
  • Which two conditions would result in the weakest electric force between objects?
    11·1 answer
  • What do we mean when we say that 50 million tons of coal could replace the use of 0.6 MBPD (million barrels per day) of oil used
    14·1 answer
  • You will be helping Galileo perform the experiment to determine if objects with different mass fall at the same, or different, r
    7·1 answer
  • If a 5 Kg ball is attached to the end of a string, and it has a velocity of 10 m/s, what is the centripetal acceleration if the
    7·1 answer
  • Why is the unit of momentum called derived unit​
    10·1 answer
  • A regular atom has a net charge of _____
    12·1 answer
  • Which quantities decrease as the distance between a planet and the Sun increases? Check all that apply.
    7·2 answers
  • . Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
    6·1 answer
  • Panel A shows a ball shortly after being thrown upward. Panel B shows the same ball in an instant on its way down. Suppose air r
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!