All categories

3 years ago

The specific heat of aluminum is 900 J/(kg·°C). How much heat does it take to raise 1 kg of aluminum 4°C?

Physics

2 answers:

8090 [49]3 years ago

4 0

<span>B. 9.04 × 102 J, hope this helps</span>

stira [4]3 years ago

3 0

The answer would be: B

You might be interested in

The solar system is 25,000 light years from the center of our Milky Way galaxy. One light year is the distance light travels in

aksik [14]

Answer:

The time period of the solar system's orbit is $2.05\times10^{8}\ year$

Explanation:

Given that,

Distance = 25000 light year

Speed of light $c=3\times10^{8}\ m/s$

Speed of astronomers = 230 m/s

Suppose the orbit is circular, we need to find the time period of the solar system's orbit.

We need to calculate the time period

Using formula of time period

$t = \dfrac{d}{v}$

$t = \dfrac{2\pi r}{v}$

r = distance

t = time

v = speed

Put the value into the formula

$t=\dfrac{2\pi\times25000\times9.46\times10^{15}}{230\times10^{3}}$

$t=6.460\times10^{15}\ sec$

Time in years

$t=\dfrac{6.460\times10^{15}}{3.15\times10^{7}}$

$t=2.05\times10^{8}\ year$

Hence, The time period of the solar system's orbit is $2.05\times10^{8}\ year$

4 0

3 years ago

| The electric field 5.0 cm from a very long charged wire is (2000 N/C, toward the wire). What is the charge (in nC) on a 1.0-cm

Serggg [28]

Answer:

The charge is 0.056 nC.

Explanation:

Given that,

Electric field = 2000 N/C

Distance = 5.0 cm

We need to calculate the charge density

Using formula of charge density

$E=\dfrac{\lambda}{2\pi\times\epsilon_{0}r}$

$\lambda=2\pi\times\epsilon_{0}\times r\times E$

Put the value into the formula

$\lambda=2\pi\times8.85\times10^{-12}\times5.0\times10^{-2}\times2000$

$\lambda=5.56\times10^{-9}\ C/m$

We need to calculate the charge in 1.0 cm

Using formula of charge

$Charge = \lambda\times\text{length of segment}$

$Charge =5.56\times10^{-9}\times1.0\times10^{-2}$

$Charge=0.056\times10^{-9}\ C$

$Charge=0.056\ nC$

Hence, The charge is 0.056 nC.

3 0

3 years ago

When Earth and the Moon are separated by a

Wewaii [24]

Using the Universal Gratitation Law, we have:

$F= \frac{MmG}{d^2} \\ MmG=2*10^{20}*(3.84*10^8)^2 \\ MmG=29.4912*10^36$

Again applying the formula in the new situation, comes:

$F= \frac{MmG}{d^2} \\ F= \frac{29.4912*10^36}{(1.92*10^8)^2} \\ \boxed {F=8*10^{20}}$

Number 4

If you notice any mistake in my english, please let me know, because i am not native.

6 0

4 years ago

Read 2 more answers

An object moves 4 meters north, then 3 meters east. What is the displacement of the object?

Lady bird [3.3K]

The displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

<h3>What is displacement?</h3>

Displacement is change in the position of an object. It can be described as the shortest distance between the initial and final position of an object.

The displacement of the object is calculated as follows;

d² = 4² + 3²

d² = 25

d = 5 m

Thus, the displacement of the object that moved 4 meters north, then 3 meters east is 5 meters.

Learn more about displacement here: brainly.com/question/2109763

#SPJ2

4 0

2 years ago

Read 2 more answers

2 PUILLS

Colt1911 [192]

B-friction acts in direction opposite to objects motion

5 0

3 years ago

Read 2 more answers