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2 years ago

Need help!!! Physics due tomorrow help, help ,help

Physics

1 answer:

kicyunya [14]2 years ago

6 0

I don't know, but for 5, to keep an object still( in zero gravity), both of the forces must be equal. Just a guess. I am in 8th grade math.

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Please answer B, C, E, and D

Natasha_Volkova [10]

B- the biosphere needs the hydrosphere (water) to survive. They drink the water.
C- the biosphere use the géosphère as land/ home.
D-The géosphère depends on the hydrosphere for water. Without the water, the géosphère would eventually turn dry and nothing will be able to grow
E- The animals in the hydrosphere need air to survive. The gills on sea animals are used to filter out the water from the oxygen and use the oxygen to breathe. As the fish opens its mouth, water runs over the gills, and blood in the capillaries picks up oxygen that's dissolved in the water.

3 0

1 year ago

Read 2 more answers

For a rigid body in rotational motion, what can be stated about the angular velocity of all of the particles? The linear velocit

Andru [333]

Answer:

Explanation:

In case of rotational motion, every particle is rotated with a same angular velocity .

The relation between linear velocity and angular velocity is

V = r w

Where v is linear velocity, r is the radius of circular path and w be the angular velocity.

Here w is same for all the particles but every particle has different radius of circular path I which they are rotating, so linear velocity is differnet for all.

7 0

2 years ago

A 2.0 kg book is lying on a 0.78 m -high table. You pick it up and place it on a bookshelf 2.1 m above the floor.Part ADuring th

zloy xaker [14]

Given data:

* The mass of the book is 2 kg.

* The initial height of the book is 0.78 m.

* The final height of the book is 2.1 m.

Solution:

(A). The work done by the gravity on the book is,

$W=mg(h_f-h_i)$

where m is the mass, g is the acceleration due to gravity, h_i is the initial height and h_f is the final height,

The work is done in moving the object in upward direction where as the gravitational force is acting in the down ward direction. Thus, the value g (acceleration due to gravity) is taken as negative in this case.

Substituting the known values,

$\begin{gathered} W=2\times(-9.8)\times(2.1-0.78) \\ W=-19.6\times1.32 \\ W=-25.9\text{ J} \end{gathered}$

Thus, the work done by the gravitational force is -25.9 J.

(B). The work done by the hand on the moving the book is,

$\begin{gathered} W_1=-W \\ =25.9\text{ J} \end{gathered}$

Thus, the work done by the hand on the book is 25.9 J.

7 0

2 months 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

2 years ago

Which of the following is a false statement? Select one:

Art [367]

Answer:

The false statement is in option 'd': The center of mass of an object must lie within the object.

Explanation:

Center of mass is a theoretical point in a system of particles where the whole mass of the system is assumed to be concentrated.

Mathematically the position vector of center of mass is defined as

$\overrightarrow{r}_{com}=\int \overrightarrow{r}_{i}dm$

where,

$\overrightarrow{r}_{i}$ is the position vector of the mass dm.

As we can see for homogenous symmetrical objects such as a sphere,cube,disc the center of mass is located at the centroid of the shapes itself but in many shapes it is located outside the body also.

Examples of shapes in which center of mass is located outside the body:

1) Horseshoe shaped body.

2) A thin ring.

In many cases we can make shapes of bodies whose center of mass lies outside the body.

6 0

2 years ago