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Anvisha [2.4K]
3 years ago
7

Show that the units of g, previously given as N/kg, are also m/s^2.

Physics
1 answer:
cupoosta [38]3 years ago
8 0

Explanation:

N/ Kg .................(1)

where,

1N= 1( Kg× m)/ s² ( sub in 1)

so,

(Kg×m)/ ( kg × s²)

kg is removed with kg

= m/ s²

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On the graph of voltage versus current, which line represents a 2.0 Ω resistor?​
Vikki [24]

Answer:

<h2>line B</h2>

Explanation:

According to ohm's law V = IR where;

V i sthe supply voltage (in volts)

I = supply current (in amperes)

R = resistance (in ohms)

In order to calculate the line that is equal to 2ohms, we need to calculate the slope of each line using the formula.

For line B, R = ΔV/ΔI

R = V₂-V₁/I₂-I₁

R = 14.0-4.0/7.0-2.0

R = 10.0/5.0

R = 2.0ohms

Since the slope of line B is equal to 2 ohms, this shows that the line B is the one that represents the 2ohms resistor.

3 0
3 years ago
Which statement about moons is true?
kotegsom [21]

Answer:

A

Explanation:

All of the other answers don't make much sense

8 0
3 years ago
A delivery truck travels with a constant velocity up an 8 slope. A 60 kg box sits on the floor of the truck and, because of sta
Blababa [14]

Explanation:

Given that,

The slope of the ramp, \theta=8^{\circ}

Mass of the box, m = 60 kg

(a) Distance covered by the truck up the slope, d = 300 m

Initially the truck moves with a constant velocity. We know that the net work done on the box is equal to 0 as per work energy theorem as :

W=\dfrac{1}{2}m(v^2-u^2)=0

u and v are the initial and the final velocity of the truck

(b) The work done on the box by the force of gravity is given by :

W=Fd\ cos\theta

Here, \theta=90+8=98^{\circ}

W=mgd\ cos\theta

W=60\times 9.8\times 300\ cos(98)

W = -24550.13 J

(c) What is the work done on the box by the normal force is equal to 0 as the angle between the force and the displacement is 90 degrees.

(d) The work done by friction is given by :

W_f=-W

W_f=24550.13\ J

Hence, this is the required solution.

6 0
3 years ago
How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the o
melisa1 [442]

Question:

A spaceship enters the solar system moving toward the Sun at a constant speed relative to the Sun. By its own clock, the time elapsed between the time it crosses the orbit of Jupiter and the time it crosses the orbit of Mars is 35.0 minutes

How fast is the spaceship traveling towards the Sun? The radius of the orbit of Jupiter is 43.2 light-minutes, and that of the orbit of Mars is 12.6 light-minutes.

Answer:

S = 5.508 × 10¹¹m

V = 2.62 × 10⁸ m/s

Explanation:

The radius of the orbit of Jupiter, Rj is 43.2 light-minutes

radius of the orbit of Mars, Rm is 12.6 light-minutes

Distance travelled S = (Rj - Rm)

= 43.2 - 12.6 = 30.6 light- minutes

= 30.6 × (3 ×10⁸m/s) × 60 s

= 5.508 × 10¹¹m

time = 35mins = (35 × 60 secs)

= 2100 secs

speed = distance/time

V = 5.508 × 10¹¹m / 2100 s

V = 2.62 × 10⁸ m/s

7 0
3 years ago
If a person walks 1km north , 5 km West , 3km south and 7km east, find the resultant displacement vector
raketka [301]

Y = +1-3 = -2  

X = -5+7 = +2  

D = √2^2-2^2 = 2√1+1 = 2√2 km

3 0
3 years ago
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