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krok68 [10]
3 years ago
12

The record blue whale shown below had a mass of 1.90 × 105 kg. Assuming that its average density was 0.900 g/cm3, as has been me

asured for other blue whales, what was the volume of the whale in cubic meters (m3)?

Physics
1 answer:
Agata [3.3K]3 years ago
8 0

Answer:

replace 920 with 900

Explanation:

You might be interested in
А A van accelerates from Amst to zomst in 8s. How far does it<br>travel in<br>this time?​
8_murik_8 [283]

Answer:

1663m

Explanation:

mark brainliest

6 0
2 years ago
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
Natali5045456 [20]

Answer:

A. 1.4 m/s to the left

Explanation:

To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:

M_{before} = M_{after}

where:

M = momentum [kg*m/s]

M = m*v

where:

m = mass [kg]

v = velocity [m/s]

(m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})

where:

m1 = mass of the basketball = 0.5 [kg]

v1 = velocity of the basketball before the collision = 5 [m/s]

m2 = mass of the tennis ball = 0.05 [kg]

v2 = velocity of the tennis ball before the collision = - 30 [m/s]

v3 =  velocity of the basketball after the collision [m/s]

v4 = velocity of the tennis ball after the collision = 34 [m/s]

Now replacing and solving:

(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)

1 - (0.05*34) = 0.5*v3

- 0.7 = 0.5*v

v = - 1.4 [m/s]

The negative sign means that the movement is towards left

3 0
2 years ago
In si units, the electric field in an electromagnetic wave is described by ey = 104 sin(1.40 107x − ωt). (a) find the amplitude
melamori03 [73]
Answers:
(a) B_o  = 0.3466μT
(b) \lambda = 0.4488μm
(c) f = 6.68 * 10^{14}Hz

Explanation:
Given electric field(in y direction) equation:
E_y = 104sin(1.40 * 10^7 x -\omega t)

(a) The amplitude of electric field is E_o = 104. Hence

The amplitude of magnetic field oscillations is B_o =  \frac{E_o}{c}
Where c = speed of light

Therefore,
B_o =  \frac{104}{3*10^8} = 0.3466μT (Where T is in seconds--signifies the oscillations)

(b) To find the wavelength use:
\frac{2 \pi }{\lambda} = 1.40 * 10^7
\lambda =  \frac{2 \pi}{1.40} * 10^{-7}
\lambda =  0.4488μm

(c) Since c = fλ
=> f = c/λ

Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f = 6.68 * 10^{14}Hz


6 0
3 years ago
A small steel roulette ball rolls around the inside of a 30 cm diameter roulette wheel. It is spun at 150 rpm, but is slows to 6
liraira [26]

Solution :

Given

Diameter of the roulette ball = 30 cm

The speed ball spun at the beginning = 150 rpm

The speed of the ball during a period of 5 seconds = 60 rpm

Therefore, change of speed in 5 seconds = 150 - 60

                                                                      = 90 rpm

Therefore,

90 revolutions in 1 minute

or In 1 minute the ball revolves 90 times

i.e. 1 min = 90 rev

     60 sec = 90 rev

        1 sec = 90/ 60 rec

         5 sec = $\frac{90}{60}\times 5$

                   = 75 rev

Therefore, the ball made 75 revolutions during the 5 seconds.

7 0
2 years ago
g An electron enters a region of space containing a uniform 1.63 × 10 − 5 T magnetic field. Its speed is 121 m/s and it enters p
kolbaska11 [484]

Answer:

i. The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

ii. The frequency 'f' of the motion is 455.44 KHz.

Explanation:

The radius 'r' of the electron's path is called a gyroradius. Gyroradius is the radius of the circular motion of a charged particle in the presence of a uniform magnetic field.

                 r = \frac{mv}{qB}

Where: B is the strength magnetic field, q is the charge, v is its velocity and m is the mass of the particle.

From the question, B = 1.63 × 10^{-5}T, v = 121 m/s, Θ = 90^{0} (since it enters perpendicularly to the field), q = e  = 1.6 × 10^{-19}C and m = 9.11 × 10^{-31}Kg.

Thus,

         r = \frac{mv}{qB} ÷ sinΘ

But,  sinΘ =  sin 90^{0} = 1.

So that;

          r = \frac{mv}{qB}

            = (9.11 × 10^{-31} × 121) ÷ (1.6 × 10^{-19}  × 1.63 × 10^{-5})

            = 1.10231 × 10^{-28}   ÷ 2.608 × 10^{-24}

            = 4.2266 × 10^{-5}

            = 4.23 × 10^{-5} m

The radius 'r' of the electron's path is 4.23 × 10^{-5} m.

B. The frequency 'f' of the motion is called cyclotron frequency;

           f = \frac{qB}{2\pi m}

             =  (1.6 × 10^{-19}  × 1.63 × 10^{-5}) ÷ (2 ×\frac{22}{7} × 9.11 × 10^{-31})

             =  2.608 × 10^{-24} ÷  5.7263 × 10^{-30}

             = 455442.4323

          f  = 455.44 KHz

The frequency 'f' of the motion is 455.44 KHz.

3 0
3 years ago
Read 2 more answers
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