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3 years ago

After their loss to U of A, the ASU sumdevils traveled back to Tempe which is 187

Physics

1 answer:

Dmitry_Shevchenko [17]3 years ago

5 0

The average speed is 20.8 m/s

Explanation:

The average speed for the trip is given by:

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

where

d is the distance covered

t is the time elapsed

For the trip in this problem, we have:

d = 187 km = 187,000 m is the distance travelled

The initial time is 10:00 pm while the arriving time is 12:30 am: this means that the time elapsed is 2.5 hours. Converting into seconds,

$t = 2.5 h \cdot (60)(60)=9000 s$

Therefore, the average speed for the trip is

$v=\frac{187,000}{9000}=20.8 m/s$

Learn more about average speed:

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Oil having a density of 930 kg/m^3 floats on

Zolol [24]

Answer:

0.0268 m

Explanation:

Draw a free body diagram of the block. There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.

Sum of forces in the y direction:

∑F = ma

B₁ + B₂ − mg = 0

ρ₁V₁g + ρ₂V₂g − mg = 0

ρ₁V₁ + ρ₂V₂ = m

ρ₁V₁ + ρ₂V₂ = ρV

ρ₁Ah₁ + ρ₂Ah₂ = ρAh

ρ₁h₁ + ρ₂h₂ = ρh

(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

Since the block is fully submerged, h₁ + h₂ = 4.93 cm.

(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

h₂ = 2.68 cm

h₂ = 0.0268 m

4 0

3 years ago

An attacker at the base of a castle wall 3.95 m high throws a rock straight up with speed 5.00 m/s from a height of 1.60 m above

s344n2d4d5 [400]

Answer:

a) No

b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.

Explanation:

Given:

Height of the wall = 3.95m

Initial height = 1.60m

Initial speed = 5.00m/s

distance between the initial height and wall top = 3.95 - 1.60 = 2.35m

Using the formula;

v^2 = u^2 + 2as ....1

Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled

From equation 1

s = (v^2 - u^2)/2a ...2

Since the rock t moving up,

the acceleration = -g = -9.8m/s2

s = maximum height travelled

v = 0 (at maximum height velocity is zero)

Substituting into equation 2

s = (0 - 5^2)/(2×-9.8) = 1.28m

Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall

b) Using equation 1:

u^2 = v^2 - 2as

v = 0

a = -9.8m/s

s = 2.35m. (distance between the initial height and wall top)

u^2 = 0 - 2(-9.8 × 2.35)

u^2 = 46.06

u = √46.06

u = 6.79m/s

Therefore, the rock must have a minimum initial speed of 6.79m/s

8 0

3 years ago

If the shoe has less mass, it will experience _______________ (more, less, the same) friction as it would with more mass.

Valentin [98]

More I’m pretty sure but no promises

4 0

3 years ago

The distance versus this plot for a particular object shows a quadratic relationship. Which column of distance data is possible

SOVA2 [1]

Answer:

Explanation:

In Column A, the pattern is 1², 2², 3², etc. So x(t) = t², which is quadratic.

In Column B, the pattern is 2*1, 2*2, 2*3, etc. So x(t) = 2t, which is linear.

In Column C, the pattern is 9*1, 9*2, 9*3, etc. So x(t) = 9t, which is linear.

In Column D, the pattern is 1/1, 1/2, 1/3, etc. So x(t) = 1/t, which is reciprocal.

In Column E, the pattern is 1/1², 1/2², 1/3², etc. So x(t) = 1/t², which is reciprocal.

5 0

4 years ago

The nuclear equation is incomplete. Superscript 239 Subscript 94 Baseline P u + Superscript 1 Subscript 0 Baseline n yields Supe

Crank

Answer:

The particle which completes the given equation is : $_{54}^{138}\textrm{Xe}$

Explanation:

The given reaction is of a fission reaction:

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_Z^A\textrm{X}+2_0^1\textrm{n}$

Total mass on the reactant side is equal to the total mass on the product side:

239 + 1 = 100 +A+ 2

A = 138

Sum of atomic numbers on the reactant side is equal to the sum of atomic number on the product side:

94 + 1(0) = 40 + Z + 2(0)

Z = 54

So atomic number 54 id of Xenon.

The particle which completes the given equation is :

$_{94}^{239}\textrm{Pu}+_0^1\textrm{n}\rightarrow _{40}^{100}\textrm{Zr}+_{54}^{138}\textrm{Xe}+2_0^1\textrm{n}$

3 0

3 years ago