All categories

3 years ago

Which of these actions would increase heat transfer between two objects?

Physics

2 answers:

xxMikexx [17]3 years ago

6 0

Answer:

Explanation:

increasing the area of their contact.

BTS ARMY

Inga [223]3 years ago

5 0

Answer: B.) increasing the area of their contact

Explanation: i hope this helps :)

You might be interested in

Seasons are more noticeable in places that are____from to the equator

Liula [17]

1. farther
2. true

the equator is a weirdo

4 0

3 years ago

HELP<br><br> These images represent waves. <br><br> Which wave has the highest frequency?

Ierofanga [76]

Wave C

because it is the closest or each wave is closer than the others

7 0

3 years ago

Read 2 more answers

A particle is confined to the x-axis between x = 0 and x = 1 nm. The potential energy U = 0 inside this region and U is infinite

DiKsa [7]

Answer:

Explanation:

According to heisenberg uncertainty Principle

Δx Δp ≥ h / 4π , where Δx is uncertainty in position , Δp is uncertainty in momentum .

Given

Δx = 1 nm

Δp ≥ h /1nm x 4π

≥ 6.6 x 10⁻³⁴ / 10⁻⁹ x 4 π

≥ . 5254 x ⁻²⁵

h / λ ≥ . 5254 x ⁻²⁵

6.6 x 10⁻³⁴ /. 5254 x ⁻²⁵ ≥ λ

12.56 x 10⁻⁹ ≥ λ

longest wave length = 12.56 n m

6 0

3 years ago

Can someone help me with this question

Anna71 [15]

Answer:

Ionic- Ion

Covalent- Molecules

Explanation:

7 0

2 years ago

To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r

sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

$V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}$

a).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{t}= 0.7 m^{2}$ , $D_{t}= 0.28$

$F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N$

$F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N$

b).

ρ $= 1.2 \frac{kg}{m^{3} }$ , $A_{h}= 2.44 m^{2}$ , $D_{h}= 0.57$

$F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N$

$F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N$

6 0

3 years ago