Which best explains why we are more aware of gravitational forces than electrical forces?

A shot-putter accelerates a 7.2 kg shot from rest to 17 m/s . what work did the shot-putter do on the ball?

garri49 [273]

<span>1.0x10^3 Joules The kinetic energy a body has is expressed as the equation E = 0.5 M V^2 where E = Energy M = Mass V = Velocity Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion E = 0.5 * 7.2 kg * (17 m/s)^2 E = 3.6 kg * 289 m^2/s^2 E = 1040.4 kg*m^2/s^2 E = 1040.4 J So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>

6 0

4 years ago

What can a human still do better and faster than any machine learning solution<br>

emmasim [6.3K]

Answer:

mistakes I guess lol don't take serious

3 0

3 years ago

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.

Lelechka [254]

Answer:

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

Explanation:

Energy conservation law: In isolated system the amount of total energy remains constant.

The types of energy are

Kinetic energy.
Potential energy.

Kinetic energy $=\frac{1}{2} mv^2$

Potential energy = $\frac{Kq_1q_2}{d}$

Here, q₁= +5.00×10⁻⁴C

q₂=-3.00×10⁻⁴C

d= distance = 4.00 m

V = velocity = 800 m/s

Total energy(E) =Kinetic energy+Potential energy

$=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}$

=(1280-337.5)J

=942.5 J

Total energy of a system remains constant.

Therefore,

E $=\frac{1}{2} mv^2$ + $\frac{Kq_1q_2}{d}$

$\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}$

$\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750$

$\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750$

$\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}$

$\Rightarrow v= 1961.19$ m/s

Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

5 0

3 years ago

A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calc

Yuki888 [10]

Answer:

The wavelength will be 4 cm, frequency will be 4.66 Hz and wave speed is 18.6 cm/sec

Explanation:

Given:

No. of crest = 13

No. of trough = 15

Time = 3 seconds

Hence, 1 crest or 1 trough = $\frac{1}{2}\lambda$

therefore,

13 C + 15 T = $28(\frac{1}{2}\lambda)$

= $14\lambda$

Time given 3 seconds

= $\frac{3}{14}s$

$\nu= \frac{14}{3}$

$\nu= 4.6 Hz \approx 5 Hz$

2 cm distance is travelled is time period

$\lambda = 4 cm$

Again wave will travel in 1 T = 4 cm

wave speed v = $\lambda\times\nu$

= $4\times\frac{14}{3}$

= 18.6 cm/s

5 0

3 years ago

(a) Calculate the distance that light travels in 0.5 seconds, and express the answer in units of the circumference of the earth.

Elan Coil [88]

Answer:

a) 3.74 circumference of earth

b)1.67× 10⁻⁸ sec

Explanation:

a) given,

time = 0.5 second

distance = speed × time

= 3 x 10⁸ × 0.5

= 1.5 × 10⁸ m

circumference of earth = 40075 km

= 4.01 × 10⁷m

distance with respect to circumference= $\dfrac{1.5 \times 10^8}{4.01 \times 10^7}$

= 3.74 circumference of earth

b) distance = 5 m

time = $\dfrac{distance}{speed}$

= $\dfrac{5}{3\times 10^8}$

= 1.67× 10⁻⁸ sec

6 0

4 years ago