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
NeTakaya
3 years ago
14

Convert 550 cm into m. Please show your work

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0
There are 100cm in one metre. Hence, divide 100 by 550 to figure out for much 550cm is in meters. It is 5.5m.
You might be interested in
If you are driving 128.4 km/h along a straight road and you look down for 3.0s, how far do you travel during this inattentive pe
ser-zykov [4K]

Answer:

107 m

Explanation:

Convert km/h to m/s:

128.4 km/h × (1000 m / km) × (1 h / 3600 s) = 35.67 m/s

Distance = rate × time

d = 35.67 m/s × 3.0 s

d = 107 m

8 0
3 years ago
If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is: A. 705°C. B. 873°C. C. 273°C. D. 327°C
zlopas [31]

Answer:

D). 327 ^0 C

Explanation:

As we know that temperature scale is linear so we will have

\frac{^0C - 0}{100 - 0} = \frac{K - 273}{373 - 273}

now we have

\frac{^0 C - 0}{100} = \frac{K - 273}{100}

so the relation between two scales is given as

^0 C = K - 273

now we know that in kelvin scale the absolute temperature is 600 K

so now we have

T = 600 - 273 = 327 ^0 C

so correct answer is

D). 327 ^0 C

4 0
3 years ago
A fireworks rocket is fired vertically upward. At its maximum height of 90.0 m , it explodes and breaks into two pieces, one wit
Alex73 [517]

Answer:

Ai. Speed of the fragment with mass mA= 1.35 kg is 34.64 m/s

Aii. Speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. 475.3 m

Explanation:

A. Determination of the speed of each fragment.

I. Determination of the speed of the fragment with mass mA = 1.35 kg

Mass of fragment (m₁) = 1.35 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₁) =?

KE = ½m₁u₁²

810 = ½ × 1.35 × u₁²

810 = 0.675 × u₁²

Divide both side by 0.675

u₁² = 810 / 0.675

u₁² = 1200

Take the square root of both side.

u₁ = √1200

u₁ = 34.64 m/s

Therefore, the speed of the fragment with mass mA = 1.35 kg is 34.64 m/s

II. I. Determination of the speed of the fragment with mass mB = 0.270 kg

Mass of fragment (m₂) = 0.270 kg

Kinetic energy (KE) = 810 J

Velocity of fragment (u₂) =?

KE = ½m₂u₂²

810 = ½ × 0.270 × u₂²

810 = 0.135 × u₂²

Divide both side by 0.135

u₂² = 810 / 0.135

u₂² = 6000

Take the square root of both side.

u₂ = √6000

u₂ = 77.46 m/s

Therefore, the speed of the fragment with mass mB = 0.270 kg is 77.46 m/s

B. Determination of the distance between the points on the ground where they land.

We'll begin by calculating the time taken for the fragments to get to the ground. This can be obtained as follow:

Maximum height (h) = 90.0 m

Acceleration due to gravity (g) = 10 m/s²

Time (t) =?

h = ½gt²

90 = ½ × 10 × t²

90 = 5 × t²

Divide both side by 5

t² = 90/5

t² = 18

Take the square root of both side

t = √18

t = 4.24 s

Thus, it will take 4.24 s for each fragments to get to the ground.

Next, we shall determine the horizontal distance travelled by the fragment with mass mA = 1.35 kg. This is illustrated below:

Velocity of fragment (u₁) = 34.64 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₁) =?

s₁ = u₁t

s₁ = 34.64 × 4.24

s₁ = 146.87 m

Next, we shall determine the horizontal distance travelled by the fragment with mass mB = 0.270 kg. This is illustrated below:

Velocity of fragment (u₂) = 77.46 m/s

Time (t) = 4.24 s

Horizontal distance travelled by the fragment (s₂) =?

s₂ = u₂t

s₂ = 77.46 × 4.24

s₂ = 328.43 m

Finally, we shall determine the distance between the points on the ground where they land.

Horizontal distance travelled by the 1st fragment (s₁) = 146.87 m

Horizontal distance travelled by the 2nd fragment (s₂) = 328.43 m

Distance apart (S) =?

S = s₁ + s₂

S = 146.87 + 328.43

S = 475.3 m

Therefore, the distance between the points on the ground where they land is 475.3 m

3 0
3 years ago
Energy needs, in total kilocalories per day, are greater during _____________ than any other time of life.
malfutka [58]
Energy needs, in total kilocalories per day, are greater during adolescence than any other time of life. The correct option among all the options that are given in the question is the second option or option "b". Only during the time of pregnancy will a girl require more kilocalories per day than during the time of adolescence. 
5 0
3 years ago
Read 2 more answers
A spacecraft orbits the Earth at height of 1600km. Calculate the escape velocity for the spacecraft.
guapka [62]

Answer:

what do u need help with

Explanation:

Calculate the escape velocity for the spacecraft. [G= 6.67×10^-11Nm^2kg^-2, mass of the Earth= 5.97×10^24kg, radius of the Earth= ...

7 0
3 years ago
Other questions:
  • 1. The experiment shown below is finished. What can you conclude from these results?
    15·1 answer
  • Work can ____ energy between objects and can cause a change in the form of energy.
    12·2 answers
  • A) An electron is moving with a speed of 3.5 x105 m/s when it encounters a magnetic field of 0.60T. The direction of the magneti
    13·1 answer
  • Two enery converrsation that take place when you warm a cup of cocoa in a microwave<br> Oven
    13·1 answer
  • What is the mass of the dog that's running 10m/s that has a momentum of 250 m/s
    5·1 answer
  • Object that is relatively close to mirror ray diagrams for convex mirrors
    15·1 answer
  • 3. Riddle:
    13·2 answers
  • San
    8·2 answers
  • Red+Blue=???????<br>Answer it.​
    13·2 answers
  • A 2300-kg car slows down at a rate of 3.0 m/s2 when approaching a stop sign. What is the magnitude of the net force causing it t
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!