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oksian1 [2.3K]
3 years ago
14

Bruce has a momentum of 430kg m/s and is running at 7.8m/s. What is Bruce's mass?

Physics
2 answers:
mixas84 [53]3 years ago
8 0

Answer:

55.13 kg

Explanation:

Momentum, P = m*v

P = m*v

m = P/v

m = 430/7.8

m = 55.13 kg

ryzh [129]3 years ago
7 0

Answer:

55.128kg

Explanation:

P= m×v

or, m=P/v = 430/7.8 = 55.128 kg

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Determine the CM of a rod assuming its linear mass density λ (its mass per unit length) varies linearly from λ = λ0 at the left
Dahasolnce [82]

Answer:

x_c= \dfrac{5}{9}L

I=\dfrac {7}{12}\lambda_ 0 L^3

Explanation:

Here mass density of rod is varying so we have to use the concept of integration to find mass and location of center of mass.

At any  distance x from point A mass density

\lambda =\lambda_0+ \dfrac{2\lambda _o-\lambda _o}{L}x

\lambda =\lambda_0+ \dfrac{\lambda _o}{L}x

Lets take element mass at distance x

dm =λ dx

mass moment of inertia

dI=\lambda x^2dx

So total moment of inertia

I=\int_{0}^{L}\lambda x^2dx

By putting the values

I=\int_{0}^{L}\lambda_ ox+ \dfrac{\lambda _o}{L}x^3 dx

By integrating above we can find that

I=\dfrac {7}{12}\lambda_ 0 L^3

Now to find location of center mass

x_c = \dfrac{\int xdm}{dm}

x_c = \dfrac{\int_{0}^{L} \lambda_ 0(1+\dfrac{x}{L})xdx}{\int_{0}^{L} \lambda_0(1+\dfrac{x}{L})}

Now by integrating the above

x_c=\dfrac{\dfrac{L^2}{2}+\dfrac{L^3}{3L}}{L+\dfrac{L^2}{2L}}

x_c= \dfrac{5}{9}L

So mass moment of inertia I=\dfrac {7}{12}\lambda_ 0 L^3 and location of center of mass  x_c= \dfrac{5}{9}L

8 0
3 years ago
A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstanc
guajiro [1.7K]

Answer:

acceleration = 0.022 m/s^2

distance = 8.3 x 10^7 m

speed = 1.9 x 10 ^3 m/s

Explanation:

the parameters given are:

mass = 900kg

force = 20N

  • from the formula force = mass x acceleration

acceleration = force / mass

acceleration = 20 / 900

acceleration = 0.022 m/s^2

  • distance travelled in 1 day (86,400 seconds) = (1/2) x a x t^2

   (1/2) x 0.022 x (86,400^2) = 8.3 x 10^7 m

  • speed of the sun yatch (v) = a x t

0.22 x 86400 = 1.9 x 10 ^3 m/s

8 0
2 years ago
Two loudspeakers emit identical sound waves along the x-axis. The sound at a point on the axis has maximum intensity when the sp
EleoNora [17]

The concept required to solve this problem is related to the wavelength.

The wavelength can be defined as the distance between two positive crests of a wave.

The waves are in phase, then the first distance is

\Delta x_1 = 20cm

And out of the phase when

\Delta x_2 = 30cm

Thus the wavelength is

\Delta x_2-\Delta x_1 = \frac{\lambda}{2}

Here,

\lambda =  Wavelength

If we rearrange the equation to find it, we will have

\lambda = 2 (\Delta x_2-\Delta x_1 )

\lambda= 2(30-20)

\lambda = 20cm

Therefore the wavelength of the sound is 20cm.

5 0
3 years ago
What is F = m x a in Newton's laws of motion?
madam [21]

Answer:

This is Newton's second law.

<u>Newton's second law text:</u>

(If a resultant force acts on a body, then an acceleration will give it an acceleration, the magnitude of which is directly proportional to the amount of the net force, and a direction is in the direction of the net force itself)

F=ma

net force = mass x acceleration

I hope I helped you^_^

8 0
2 years ago
What is the electric potential of a 2.2 µC charge at a distance of 6.3 m from the charge? Recall that Coulomb’s constant is k =
iren2701 [21]
1) The electric potential at a distance r from a single point charge is given by
V(r) = k  \frac{q}{r}
where k is the Coulomb's constant, q is the charge and r is the distance from the charge.

The charge in this problem is
q=2.2 \mu C =2.2 \cdot 10^{-6} C
So the potential at distance r=6.3 m is
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2 C^{-2})  \frac{2.2 \cdot 10^{-6} C}{6.3 m}=3139 V

2) By using the same formula as before, we can find the electric potential at distance r=99 m from the charge:
V=k \frac{q}{r}=(8.99 \cdot 10^9 Nm^2C^{-2})  \frac{2.2 \cdot 10^{-6} C}{99 m}=198 V
3 0
3 years ago
Read 2 more answers
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