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Elanso [62]
2 years ago
13

What is the KE of a bicycle with a mass of 41kg traveling at a velocity of 8 m/s ?

Physics
1 answer:
zhuklara [117]2 years ago
3 0

Answer:

What is the kinetic energy of a bicycle with a mass of 14 kg traveling at a velocity of 3 m/s? Physics ... K=12mv2=12×14 kg × (3 m/s)2=63 Joule.

1 answer  

63 Joule Explanation: K=12mv2=12×14 kg × (3 m/s)2=63 Joule

Explanation:

K=12mv2=12×14 kg × (3 m/s)2=63 Joule.

63 Joule Explanation: K=12mv2=12×14 kg × (3 m/s)2=63 Joule

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Gold has a density of 19.3 g/cm3. What is the mass of a 5 cm3 block of gold?
Tems11 [23]
Do not forget that mass = <span>volume x density
</span>Mass of 1 cm^3 = Density[/tex]
mass of 2 cm^3 = 19.3 g + 19.3 g = 2*19.3 g
Then eventually we can find <span>mass of 5 cm^3 : = 
</span>19.3 g + 19.3 g+19.3g+19.3g+19.3g= 5*19.3 g
So the answer is D
<span>And that's it. I'm sure it will help.</span>
5 0
3 years ago
A projectile lands at the same height from which it was launched. which initial velocity will result
Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

5 0
3 years ago
A force of 265.1 N acts on an object to produce an acceleration of 14.52 m/s^2. What is the mass of the object?
astra-53 [7]

The answer is :


18.26


Hope I helped.

6 0
3 years ago
Read 2 more answers
Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir
strojnjashka [21]

Answer:

1.475\times 10^{-13}\ C/m^3

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area

h = Altitude = 600 m

Electric flux through the top would be

-110A (negative as the electric field is going into the volume)

At the bottom

120A

Total flux through the volume

\phi=120-110\\\Rightarrow \phi=10A

Electric flux is given by

\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0

Charge per volume is given by

\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3

The volume charge density is 1.475\times 10^{-13}\ C/m^3

7 0
3 years ago
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
JulijaS [17]

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

5 0
3 years ago
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