Eddie and Val observed the picture of an athlete running in a race.

Solving:

STatiana [176]

Answer:

1. 31,536,000 seconds

2. Car B traveled a longer distance

3. Volume of box = 0.235887 cubic meters

Explanation:

Q1. Age in seconds

1 year = 365 days
1 day = 24 hours
1 hour = 60 minutes
1 minute = 60 seconds

Therefore 1 year = 365 x 24 x 60 x 60 = 31,536,000 seconds

In scientific notation this would be $3.1536\times 10^7 \text{ seconds}$

Q2. Comparing km and miles

Given:
1 km = 1000m and 1 m = 3.3ft, I km = 1000 x 3.3 = 3300 ft.

Convert Car A distance of 25.7km to feet :
25.7 km. = 25. 7 x 3300 ft. = 84,810 ft.

For Car B that traveled 20 miles,
20 miles = 20 x 5280 = 105,600 ft.

Since 105,600 > 84,810, car B traveled a longer distance

Q3. Volume of wooden box

The wooden box is in the shape of a rectangular prism
It volume is L x W x H
Volume = 1.525 x 0.30 x 0.5156 = 0.235887 cubic meters

5 0

9 months ago

A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is i

timurjin [86]

Answer:

1) 883 kgm2

2) 532 kgm2

3) 2.99 rad/s

4) 944 J

5) 6.87 m/s2

6) 1.8 rad/s

Explanation:

1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:

$I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2$

If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:

$I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2$

2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):

$I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2$

3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:

$I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}$

$\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}$

$\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s$

4)Kinetic energy before:

$E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J$

Kinetic energy after:

$E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J$

So the change in kinetic energy is: 2374 - 1430 = 944 J

5) $a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2$

6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.

3 0

3 years ago

The asteroid, Ida, has a small moon, Dactyl, that orbits at a speed of 5.66 m/s in an orbit of radius 90, 000m . What is Ida's m

cricket20 [7]

Answer:

4.3 x 10^16 kg

Explanation:

M = rv^2/G =[90,000 x 5.66^2] / [6.67 x 10^-11]

M = 43,226,446,776,611,694 = 4.3 x 10^16 kg - Ida's mass.

4 0

3 years ago

Two spheres are cut from a certain uniform rock. One has radius 4.10 cm. The mass of the other is eight times greater.

Misha Larkins [42]

Answer:

8.2 cm

Explanation:

mass of one sphere = m

mass of other sphere = 8 m

Radius of first sphere = r = 4.10 cm

let the radius of another sphere is R

As we know that mass = volume x density

Let the density of the rock is d.

density of rock remains same.

density of small sphere = density of big sphere

mass of small sphere / volume of small sphere =

mass of big sphere / volume of big sphere

$\frac{m}{\frac{4}{3}\pi r^{3}}=\frac{8m}{\frac{4}{3}\pi R^{3}}$

$R^{3}=8r^{3}$

R = 2 r

R = 2 x 4.10 = 8.2 cm

Thus, the radius of big sphere is 8.20 cm.

7 0

2 years ago

The diagram below shows a filament lamp rated 6.0V connected to the output of a step down transformer with a 240V input supply.

koban [17]

Guess what this is bee vs??? Is my answer to your question

7 0

2 years ago