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3 years ago

Explain how the specific heat of water helps keep our bodies from overheating

1 answer:

5 0

This high specific heat of water prevents abrupt changes in temperature and helps the body maintain temperatures that are relatively constant.

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A plane is flying east when it drops some supplies to a designated target below. The supplies land after falling for 10 seconds.

Diano4ka-milaya [45]

solution:

As Given plane is flying in east direction.

It throws back some supplies to designated target.

Time taken by the supply to reach the target =10 seconds

g = Acceleration due to gravity = - 9.8 m/s²[Taken negative as object is falling Downwards]

As we have to find distance from the ground to plane which is given by d.

d = $\frac{1}{2}\times g\times t^2$

= $\frac{1}{2}\times (9.8) \times(100) =50\times 9.8=490$ meters

Distance from the ground where supplies has to be land to plane = Option B =490 meters

6 0

3 years ago

Read 2 more answers

You are on an airplane traveling 30° south of due west at 180 m/s with respect to the air. The air is moving with a speed 31 m/s

BlackZzzverrR [31]

1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>

6 0

3 years ago

A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra

kkurt [141]

Answer:

$V_{ft}= 317 cm/s$

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

$P_i = P_f$

Where:

$P_i=M_cV_{ic} + M_tV_{it}$

$P_f = M_cV_{fc} + M_tV_{ft}$

Now:

$M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}$

Where $M_c$ is the mass of the car, $V_{ic}$ is the initial velocity of the car, $M_t$ is the mass of train, $V_{fc}$ is the final velocity of the car and $V_{ft}$ is the final velocity of the train.

Replacing data:

$(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}$

Solving for $V_{ft}$ :

$V_{ft}= 3.17 m/s$

Changed to cm/s, we get:

$V_{ft}= 3.17*100 = 317 cm/s$

b) The kinetic energy K is calculated as:

K = $\frac{1}{2}MV^2$

where M is the mass and V is the velocity.

So, the initial K is:

$K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2$

$K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2$

$K_i = 22.06 J$

And the final K is:

$K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2$

$K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2$

$K_f = 19.61 J$

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0

3 years ago

A 2-kg box sits on a horizontal table. the force of friction between the box and the table is 10 n. the box is pushed to the rig

dangina [55]

By Newton's 2nd law of motion, F = ma, where F is force, m is mass, and a is acceleration.

Rearranging this equation to find acceleration would give us:
a = F/m

The horizontal force to the right is 10N, because the box is pushed to the right with a force of 20N, and the friction force of 10N opposes that, so:
20N - 10N = 10N

The mass is 2kg.

Putting these values into the equation gives us:
a = F/m
= 10/2
= 5ms^-2

The acceleration of the box is 5ms^-2

6 0

3 years ago

What is same about all myriapods

seraphim [82]

They all have segmented limbs, a hard exoskeleton, a pair of antennae and a segmented body.

4 0

3 years ago