What happen to the salt and the sand when water is added and the mixtures is stirred?

Which one of the following types of training involves 10 to 30 minutes of high-intensity exercise?

Goshia [24]

Correct answer choice is:

C. Medium range

Explanation:

Medium range exercises are used to gain extra strength and fitness. Usually, heavyweights are used with less number of repetitions. These sort of exercises are mostly the hardest t do. All you need is to have a high level of motivation and stamina, which can be gained by running or cycling.

5 0

3 years ago

Read 2 more answers

The maximum static friction between a box and the table is 15.0N when the normal force on the box is 35.0N. What is the coeffici

Marysya12 [62]

Answer:

you are going through this week of x factor of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal

Explanation:

you are going through this week of x factor of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and technology practical pdf the new modal and F kisna of x and

6 0

2 years ago

What is the Sl unit for momentum? O kg• m O kg O kg• m O kg 2 m

Andrej [43]

Explanation:

<h2>The S. I. unit of momentum is Kg. m/sec</h2>

hope it helps you 

8 0

3 years ago

A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

3 years ago