All categories

2 years ago

A person drives 70 km/h in 1 hour to the east, then 80 km/h for another hour to the east. What

Physics

1 answer:

andre [41]2 years ago

6 0

Answer: The average velocity is 150 km/h

Explanation: 70+80=150

You might be interested in

Determine the torque<br> produced by a perpendicular force of 75<br> N at the end of a 0.2 m wrench.

trasher [3.6K]

Answer:c

Explanation:

Because I took uty

7 0

2 years ago

Read 2 more answers

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several

mamaluj [8]

Answer:

The charge on the dust particle is $q_d = 6.94 *10^{-13} \ C$

Explanation:

From the question we are told that

The length is $l = 2.0 \ m$

The width is $w = 4.0 \ m$

The charge is $q = -10\mu C= -10*10^{-6} \ C$

The mass suspended in mid-air is $m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg$

Generally the electric field on the carpet is mathematically represented as

$E = \frac{q}{ 2 * A * \epsilon _o}$

Where $\epsilon _o$ is the permittivity of free space with value $\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2$

substituting values

$E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}$

$E = -70621.5 \ N/C$

Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles

$F__{E}} = F__{G}}$

=> $q_d * E = m * g$

=> $q_d = \frac{m * g}{E}$

=> $q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}$

=> $q_d = 6.94 *10^{-13} \ C$

4 0

2 years ago

Out of aluminum,copper,steel,and glass.Which material do you think will be the best thermal conductor?

Nastasia [14]

I believe it is copper

6 0

2 years ago

Which factors affect heat transfer between a warm and a cool substance?

andrey2020 [161]

Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

D. the area of the substances

(since everything else in each of those answers are the same)

Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).

5 0

2 years ago

(b) If TH = 500°C, TC = 20°C, and Wcycle = 200 kJ, what are QH and QC, each in kJ?

BigorU [14]

Answer:

QC = 122 KJ

QH = 2.64 x 122 = 322 KJ

Explanation:

TH = 500 Degree C = 500 + 273 = 773 K

TC = 20 degree C = 20 + 273 = 293 K

W cycle = 200 KJ

Use the formula for the work done in a cycle

Wcycle = QH - QC

200 = QH - QC ..... (1)

Usse

TH / TC = QH / QC

773 / 293 = QH / QC

QH / QC = 2.64

QH = 2.64 QC Put it in equation (1)

200 = 2.64 QC - QC

QC = 122 KJ

So, QH = 2.64 x 122 = 322 KJ

4 0

2 years ago