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butalik [34]
3 years ago
12

Describe the motion for each segment below. Include start position, relative speed, and direction. Then calculate the speed of e

ach section. Finally, write the equation of the line segment for segment C.
Physics
1 answer:
ratelena [41]3 years ago
3 0
Where is the picture?
You might be interested in
Find the energy in Joules required to lift a 55.0 Megagram object a distance of 500 cm.
fredd [130]

Energy to lift something =

               (mass of the object) x (gravity) x (height of the lift).

BUT ...

This simple formula only works if you use the right units.

Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters

For this question . . .

Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms

Gravity (on Earth) = 9.8 m/second²

Height = 500 cm  =  5.0 meters

So we have ...

Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)

            =  2,696,925 joules .

That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.

The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.

7 0
3 years ago
If you weigh 690 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an
lana [24]

Answer:

Explanation:

Given that,

Mass of star M(star) = 1.99×10^30kg

Gravitational constant G

G = 6.67×10^−11 N⋅m²/kg²

Diameter d = 25km

d = 25,000m

R = d/2 = 25,000/2

R = 12,500m

Weight w = 690N

Then, the person mass which is constant can be determined using

W =mg

m = W/g

m = 690/9.81

m = 70.34kg

The acceleration due to gravity on the surface of the neutron star is can be determined using

g(star) = GM(star)/R²

g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²

g (star) = 8.49 × 10¹¹ m/s²

Then, the person weight on neutron star is

W = mg

Mass is constant, m = 70.34kg

W = 70.34 × 8.49 × 10¹¹

W = 5.98 × 10¹³ N

The weight of the person on neutron star is 5.98 × 10¹³ N

5 0
3 years ago
An igneous intrusion whose boundaries lie parallel to the layering in the surrounding country rock is considered to be:
ikadub [295]
<span>...a concordant intrusion. In geology, "concordant" means the same as "sill" -- or, an intrusion that has gotten in between older layers of rock (or even beds of volcanic lava). An intrusion with boundaries parallel to layering in surrounding rocks suggests this, meaning it is considered to be a concordant intrusion.</span>
7 0
4 years ago
7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the ca
Misha Larkins [42]

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

8 0
3 years ago
An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f
hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If  Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass m_1 = 30 kg

mass m_2 = 50 kg

\mu = 0.1

From the question; we can arrive at two cases;

That :

m_{2} a _ \ {net} }= m_2g - T   ----- equation (1)

m_{1} a _ \ {net} }=  T - mg sin \theta  - F ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a =[ 50  - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]g

80 a = 32. 4 × 10 m/s ²  (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0
3 years ago
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