A hiker carries a 20 kg backpack up a trail. After 30 minutes he is 300 m higher than his starting point.

A brick in the shape of a cube with sides 10 cm has a density of 2500 kg/m^3. What is its weight? a.) 250 N

Arisa [49]

Answer:

c.) 25 N

Explanation:

We find the volume of the brick, knowing that the volume of a cube is given by the formula:

$l=0,1 metros \\V=l^{3}\\V=(0,1\: \: m)^{3}=0,001\: \: m^{3}$

being l the side of the cube, which in this case is 10 cm or 0,1 m. Now we find the mass of the object, knowing the density and the Volume of the cube:

$m=V*d\\m=(0,001 \:\:m^{3})(2500 \: \: \frac{Kg}{m^{3}})=2,5 \:\: Kg$

We find the weight by multiplying the mass of the object with the gravity constant.

$W=m*g=(2.5 \:Kg)*(9,81 \:m/s^{2} )=24,5 N\approx25\: N$

8 0

3 years ago

After a laser bean passes through two thin parallel slits, thefirst completely dark fringes occur at ± 15.00with the original di

PSYCHO15rus [73]

Answer:

143 °

Explanation:

a ) If d be the distance between slits , λ be wavelength of light used and at angle θ nth dark fringe is formed then

d sinθ = ( 2n+1) λ/2

for first dark fringe

d sinθ = λ/2

d /λ = 1/ 2 sinθ

1 / 2 sin15

= 1.93

b )

For intensity of fringe at angle θ, the relation is

I = I₀ cos²θ

I / I₀ = cos²θ/2

Given I / I₀ =0. 1

0.1 = cos²θ/2

θ/2 = 71.5

θ = 143 °

4 0

2 years ago

If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____

Tju [1.3M]

Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)∧(T/t).
Where N ⇒ original mass of cobalt
 n ⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567

t = 3÷0.567
= 5.290626524

the half-life of Cobalt-60 is 5.29 years.

8 0

2 years ago

A system consists of electrons and protons only. It contains 150 electrons and has a total charge of +22e. What is the mass of t

liubo4ka [24]

According to the statements the number of electrons is 150, then

e = 150

But there is a positive charge of +22e, then the number of protons would be

p = 150+172

If the mass of the electrons is

$m_e = 9.11*10^{-31} kg$