What evidence supports a scientist's conclusion that fossil B is older than fossil A?

How many units measure one wavelength? a 16 b 8 c 2 d 4

pochemuha

B it makes more sense

8 0

3 years ago

A car of mass 1167 kg accelerates on a flat highway from 10 m/s to 28.0 m/s. How much work does the car's engine do on the car?

denis23 [38]

Answer:

Workdone = 465766038 Joules.

Explanation:

Given the following data;

Mass = 1167

Initial velocity = 10m/s

Final velocity =28m/s

To find the workdone;

We know that from the workdone theorem, the workdone by an object or a body is directly proportional to the kinetic energy possessed by the object due to its motion.

Mathematically, it is given by the equation;

W = Kf - Ki

W = ½MVf² - ½MVi²

Substituting into the equation

W = ½(1167)*28² - ½(1167)*10²

W = ½ * 1361889* 784 - ½ * 1361889 * 100

W = 533860488 - 68094450

Workdone = 465766038 Joules.

7 0

2 years ago

Which is not expecting acceleration?

Kamila [148]

1. Answer: A skydiver whose air resistance is equal to that of her weight.

A skydiver free falls under gravity but her rate of fall slows down due to drag -air resistance. when this air resistance becomes equal to her weight, the two get balanced and the body does not accelerate or decelerates.

2. Answer: Gravity

Contact forces are those which act when there is physical contact between two bodies. For example: normal force, tension and spring force.

Non-contact forces act between two bodies even when they are at a distance apart. For example: gravity, electric force, magnetic force etc.

3. Answer: The tendency of an object's motion to remain the same.

Inertia is a property of matter by virtue of which it tends to remain in its state of motion or rest. It does depend on mass of the object, more the mass, more is inertia. For example, cycle can be easily moved but we need real push hard for a car to move.

4. Answer: 254 N

The man pushes the box with 310 N force at an angle of 55 degrees to the horizontal.

we can write this in terms of horizontal ( $F cos \theta$ )and vertical component ( $F sin\theta$ ).

Horizontal component: $F_H=310 \times cos (55^o)= 178 N$

Vertical component: $F_v=310 \times sin (55^o)= 254 N$

The vertical component would act towards the floor making the job more difficult to move the job.

7 0

3 years ago

The pressure drop needed to force water through a horizontal 1-in diameter pipe if 0.60 psi for every 12-ft length of pipe. Dete

oksian1 [2.3K]

Answer:

The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²

The shear stress at a distance 0.5-in away from the pipe wall is 0

Explanation:

Given;

pressure drop per unit length of pipe = 0.6 psi/ft

length of the pipe = 12 feet

diameter of the pipe = 1 -in

Pressure drop per unit length in a circular pipe is given as;

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\$

make shear stress (τ) the subject of the formula

$\frac{\delta P}{L} = \frac{2 \tau}{r} \\\\\tau = \frac{\delta P *r}{2L}$

Where;

τ is the shear stress on the pipe wall.

ΔP is the pressure drop

L is the length of the pipe

r is the distance from the pipe wall

Part (a) shear stress at a distance of 0.3-in away from the pipe wall

Radius of the pipe = 0.5 -in

r = 0.5 - 0.3 = 0.2-in = 0.0167 ft

ΔP = 0.6 psi/ft

ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0.0167}{2*12} =0.06012 \ lb/ft^2$

Part (b) shear stress at a distance of 0.5-in away from the pipe wall

r = 0.5 - 0.5 = 0

$\tau = \frac{\delta P *r}{2L} = \frac{86.4 *0}{2*12} =0$

3 0

3 years ago

I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a

attashe74 [19]

The angular momentum of a particle in orbit is

l = m v r

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"

m_1 v_1 r_1 = m_2 v_2 r_2

Assuming that the mass did not change, conservation of angular momentum demands that

v_1 r_1 = v_2 r_2

or

v1 = v_2 (r_2/r_1)

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, the orbital speed of this material when it was 40,000 AU from the sun is 4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

3 0

2 years ago