All categories

3 years ago

Which of the below statements are true?

Physics

1 answer:

andrezito [222]3 years ago

4 0

Answer:

Explanation:

Displacement is the distance from the start point to the endpoint, displacement disregard the path taken or the amount traveled.

if you start at point A, then go to point B, and back to point A, the displacement is zero because you started and ended at the same point.

for this question, pretend you started at point A, went east 20 km to point B, and then west 8 km to point C, your displacement is 12 km. 12 km is the distance between point A and point C.

You might be interested in

Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0

3 years ago

Read 2 more answers

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

The most precise measurement of the speed of light in vacuum is 299,792,458 m/s. Which of the following values would you use to

Luda [366]

Answer:

v = 3.00 x 10⁸ m/s

Explanation:

given,

speed of light in vacuum = 299,792,458 m/s

speed of light in scientific notation to three significant figures

v = 2.99792458 x 10⁸ m/s

by rounding off the speed to three significant figure.

v = 3.00 x 10⁸ m/s

On the fourth place the value is greater than 5 so, on the third place 1 will be added.

now, the speed with three significant figure comes out to be

v = 3.00 x 10⁸ m/s

3 0

3 years ago

Encuentre en km.h la velocidad de un tigre que corre 550 km.h en 69min

Len [333]

Answer:

v = 478.26 km/h

Explanation:

The question is "find in km.h the speed of a tiger that runs 550 km in 69min"

Distance, d = 550 km

Time, t = 69 min = 1.15 h

We need to find the speed of the tiger. The speed of an object is equal to the total distance covered divided by time taken. So,

$v=\dfrac{550\ km}{1.15\ h}\\\\v=478.26\ km/h$

So, the speed of the tiger is 478.26 km/h.

8 0

3 years ago

A 58.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 140 m/s from the top of a cliff

strojnjashka [21]

(a) $6.43\cdot 10^5 J$

The total mechanical energy of the projectile at the beginning is the sum of the initial kinetic energy (K) and potential energy (U):

$E=K+U$

The initial kinetic energy is:

$K=\frac{1}{2}mv^2$

where m = 58.0 kg is the mass of the projectile and $v=140 m/s$ is the initial speed. Substituting,

$K=\frac{1}{2}(58 kg)(140 m/s)^2=5.68\cdot 10^5 J$

The initial potential energy is given by

$U=mgh$

where g=9.8 m/s^2 is the gravitational acceleration and h=132 m is the height of the cliff. Substituting,

$U=(58.0 kg)(9.8 m/s^2)(132 m)=7.5\cdot 10^4 J$

So, the initial mechanical energy is

$E=K+U=5.68\cdot 10^5 J+7.5\cdot 10^4 J=6.43\cdot 10^5 J$

(b) $-1.67 \cdot 10^5 J$

We need to calculate the total mechanical energy of the projectile when it reaches its maximum height of y=336 m, where it is travelling at a speed of v=99.2 m/s.

The kinetic energy is

$K=\frac{1}{2}(58 kg)(99.2 m/s)^2=2.85\cdot 10^5 J$