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agasfer [191]
3 years ago
5

Calculate the density of an object with a volume of 18m3 and a mass of 3.5kg?

Physics
1 answer:
jasenka [17]3 years ago
3 0
0.19
Explanation:you will divide 3.5 and 18
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Name an element in the same period as rubidium
Artyom0805 [142]

Answer: Potassium(K)

Explanation:

its an alkali metal placed under sodium and its over rubidium, its also the first element of period 4

5 0
2 years ago
Brycen can cover half a basketball court (about 14 meters) in 4.0 seconds flat! How fast can he run?
Vedmedyk [2.9K]

Speed is equal to distance traveled divided by the time. So it's 3.5 m/s

6 0
3 years ago
A bullet leaves a gun with a horizontal velocity of 100 m/s. Calculate the distance it would travel in 1.3 seconds.
stiks02 [169]

Ignoring air resistance, the bullet's horizontal velocity is constant:

v_x=v_{0x}=100\,\dfrac{\mathrm m}{\mathrm s}

In 1.3 seconds, we can expect it to travel

v_xt=\left(100\,\dfrac{\mathrm m}{\mathrm s}\right)(1.3\,\mathrm s)=130\,\mathrm m

4 0
3 years ago
Air (14.5 lb) undergoes a polytropic process in a closed system from p1 = 80 lbf/in2, υ1 = 4 ft3/lb to a final state where p2 =
Yanka [14]
The energy transfer in terms of work has the equation:

W = mΔ(PV)

To be consistent with units, let's convert them first as follows:

P₁ = 80 lbf/in² * (1 ft/12 in)² = 5/9 lbf/ft²
P₂ = 20 lbf/in² * (1 ft/12 in)² = 5/36 lbf/ft²
V₁ = 4 ft³/lbm
V₂ = 11 ft³/lbm

W = m(P₂V₂ - P₁V₁)
W = (14.5 lbm)[(5/36 lbf/ft²)(4 ft³/lbm) - (5/9 lbf/ft²)(11 lbm/ft³)]
W = -80.556 ft·lbf

In 1 Btu, there is 779 ft·lbf. Thus, work in Btu is:
W = -80.556 ft·lbf(1 Btu/779 ft·lbf)
<em>W = -0.1034 BTU</em>


4 0
3 years ago
A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0
3 years ago
Read 2 more answers
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