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3 years ago

What conditions are required for a lunar eclipse?

1 answer:

6 0

For a total lunar eclipse, the sun, earth and the moon must be in align with earth in between the sun and the moon.
Hope this helps.

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If the tension of the cable is 25.0 N what is the mass of the ball

marysya [2.9K]

Answer:576

Explanation:

5 0

3 years ago

Consider a large tank holding 1000 L of pure water into which a brine solution of salt begins to owat a constant rate of 6L/min.

lbvjy [14]

Answer:

T = 693.147 minutes

Explanation:

The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.

Therefore, the total salt in the tank at time t = 1000x kg

Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L

Hence, the amount of salt that is added to the tank per minute = $(6\times0.1)kg/min=0.6kg/min$

Also, there is a continuous outflow from the tank at a rate of 6 L/min.

Hence, amount of salt subtracted from the tank per minute = 6x kg/min

Now, the rate of change of salt concentration in the tank = $\frac{dx}{dt}$

So, the rate of change of salt in the tank can be given by the following equation,

$1000\frac{dx}{dt} =0.6-6x$

or, $\int\limits^{0.05}_0 {\frac{1000}{0.6-6x} } \, dx =\int\limits^T_0 {} \, dt$

or, T = 693.147 min (time taken for the tank to reach a salt concentration

of 0.05 kg/L)

3 0

3 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.50 mm above the point where the putty leaves your hand. The

LuckyWell [14K]

Answer: $4.65\ m/s$

Explanation:

Given

Distance putty has to travel is 3.5 m

The initial speed of putty is 9.50 m/s

Using equation of motion to determine the velocity of putty just before it hits ceiling

$v^2-u^2=2as$

$\Rightarrow v^2-(9.5)^2=2(-9.8)(3.5)\\\\\Rightarrow v^2=9.5^2-68.6\\\Rightarrow v=\sqrt{90.25-68.6}\\\Rightarrow v=4.65\ m/s$

So, the velocity of putty just before hitting is $4.65\ m/s$

5 0

3 years ago

Suppose that you are standing on a train accelerating at 0.20g. What minimum coefficient of static friction must exist between y

Ilia_Sergeevich [38]

Acceleration = (0.2 x g) = 1.96m/sec^2.
Accelerating force on 1kg. = (ma) = 1.96N. 
1kg. has a weight (normal force) of 9.8N. 
Coefficient µ = 1.96/9.8 = 0.2 minimum. 

Coefficient is a ratio, so holds true for any value of mass to find accelerating force acting. 
e.g. 75kg = (75 x g) = 735N. 
Accelerating force = (735 x 0.2) = 147N

5 0

4 years ago

Read 2 more answers

Two vectors A and B are at right angles to each other. The magnitude of A is 3.00. What should be the length of B, so that the m

AfilCa [17]

Answer: length of B =4.00

Explanation:

for the vectors A and B and the angle between them as x.

Magnitude of the sum of A and B is given as = √(A²+B²+2ABcosx

where

Magnitude of A = 3.00

Magnitude of the sum of A and B is 5.00

5.00=√(A²+B²+2ABcos90°

5.00= √3² +b² +0

5²= 3² +b²

25=9+b²

b²= 25-9

b² = 16

b= √16

b= 4

7 0

3 years ago