D.
Have a longer revolution time since they definitely do not get warmer, They do not have fewer moons (Jupiter has about 100 and earth has 1) they are not smaller in diameter (Earth v Jupiter)

6 0

2 years ago

A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at

Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0

3 years ago

Raw sewage is raised vertically by 5.49m in the amountof

uranmaximum [27]

Answer:

P = 1235.7646 W

Explanation:

Given data:

height of raised sewage = 5.49 m

rate of sewage =1.89*10^6 lt/day

density of sewage = 1.050 kg/m^3

power is written as

$P = \frac{work}{time}$

work = m g h

$m - mass = ( \rho V)$

h -height

mass lifted per day $= ( 1.89*10^6)(1.050)$

= 1984500 kg

time = 24 hours* 3600 seconds per hour

power $P = \frac{( 1984500)(9.8)( 5.49)}{(24)(3600)}$

P = 1235.7646 W

5 0

3 years ago

Leoni is participating in four drawing competitions. If the probability of her losing any

Alex17521 [72]

Answer:

(a) $Probability = 0.7599$

(b) $Probability = 0.2646$

Explanation:

Represent losing with L and winning with W.

So:

$L = 0.7$ --- Given

$n = 4$

Probability of winning would be:

$W = 1 - L$

$W = 1 - 0.7$

$W = 0.3$

The question illustrates binomial probability and will be solved using the following binomial expansion;

$(L + W)^4 = L^4 + 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

So:

Solving (a): Winning at least 1

We look at the above and we list out the terms where the powers of W is at least 1; i.e., 1,2,3 and 4

So, we have:

$Probability = 4L^3W + 6L^2W^2 + 4LW^3 + W^4$

Substitute value for W and L

$Probability = 4 * 0.7^3*0.3 + 6*0.7^2*0.3^2 + 4*0.7*0.3^3 + 0.3^4$

$Probability = 0.7599$

<em>Hence, the probability of her winning at least one is 0.7599</em>

Solving (a): Wining exactly 2

We look at the above and we list out the terms where the powers of W is exactly 2

So, we have:

$Probability = 6L^2W^2$

Substitute value for W and L

$Probability = 6*0.7^2*0.3^2$

$Probability = 0.2646$

<em>Hence, the probability of her winning exactly two is 0.2646</em>

6 0

2 years ago