Are the cells in this image prokaryotic or eukaryotic?

If you change the mass of a moving object, you change its _______.

Paul [167]

Momentum and inertia, Momentum=P. P=MV. M=mass and V=velocity. Mass is related to inertia so inertia=F. F=MA. A=acceleration.

6 0

4 years ago

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A rock is thrown upward from a bridge into a river below. The function f(t)=−16t2+44t+88 determines the height of the rock above

babymother [125]

1) 88 ft

2) 4.09 s

3) 1.38 s

4) 118.2 m

Explanation:

1)

For an object thrown upward and subjected to free fall, the height of the object at any time t is given by the suvat equation:

$h(t) = h_0 + ut - \frac{1}{2}gt^2$ (1)

where

$h_0$ is the height at time t = 0

$u$ is the initial vertical velocity

$g=32 ft/s^2$ is the acceleration due to gravity

The function that describes the height of the rock above the surface at a time t in this problem is

$f(t)=-16t^2+44t+88$ (2)

By comparing the terms with same degree of eq(1) and eq(2), we observe that

$h_0 = 88 ft$

which means that the rock is at height h = 88 ft when t = 0: therefore, this means that the height of the bridge above the water is 88 feet.

2)

The rock will hit the water when its height becomes zero, so when

$f(t)=0$

which means when

$0=-16t^2+44t+88$

First of all, we can simplify the equation by dividing each term by 4:

$0=-4t^2+11t+22$

This is a second-order equation, so we solve it using the usual formula and we find:

$t_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-11\pm \sqrt{(11)^2-4(-4)(22)}}{2(-4)}=\frac{-11\pm \sqrt{121+352}}{-8}=\frac{-11\pm 21.75}{-8}$

Which gives only one positive solution (we neglect the negative solution since it has no physical meaning):

t = 4.09 s

So, the rock hits the water after 4.09 seconds.

3)

Here we want to find how many seconds after being thrown does the rock reach its maximum height above the water.

For an object in free fall motion, the vertical velocity is given by the expression

$v=u-gt$

where

u is the initial velocity

g is the acceleration due to gravity

t is the time

The object reaches its maximum height when its velocity changes direction, so when the vertical velocity is zero:

$v=0$

which means

$0=u-gt$

Here we have

$u=+44 ft/s$ (initial velocity)

$g=32 ft/s^2$ (acceleration due to gravity)

Solving for t, we find the time at which this occurs:

$t=\frac{u}{g}=\frac{44}{32}=1.38 s$

4)

The maximum height of the rock can be calculated by evaluating f(t) at the time the rock reaches the maximum height, so when

t = 1.38 s

The expression that gives the height of the rock at time t is

$f(t)=-16t^2+44t+88$

Substituting t = 1.38 s, we find:

$f(1.38)=-16(1.38)^2 + 44(1.38)+88=118.2 m$

So, the maximum height reached by the rock during its motion is

$h_{max}=118.2 m$

Which means 118.2 m above the water.

3 0

4 years ago

I have to write something here, so like hello and please help

ExtremeBDS [4]

Answer:

the extension would be less the new extension might be 3 cm

Explanation:

5 0

3 years ago

A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a cons

Pie

Answer:

0.128 s

Explanation:

We have to start by calculating the net force acting on the log. We have two forces:

- The constant pulling force, forward, of F = 2500 N

- The frictional force, backward

The frictional force is given by

$F_f = \mu mg$

where

$\mu=0.45$ is the coefficient of friction

m = 300 kg is the mass of the log

$g=9.8 m/s^2$ is the acceleration of gravity

Substituting,

$F_f = (0.45)(300)(9.8)=1323 N$

So the net force acting on the log is

$F=2500 - 1323=1177 N$

Now, we can find the acceleration of the log by using Newton's second law

$F=ma$

where a is the acceleration. Re-arranging for a,

$a=\frac{F}{m}=\frac{1177}{300}=3.92 m/s^2$

And finally we can find the time it takes for the log to reach a speed of

v = 0.5 m/s

by using the suvat equation:

$v=u+at$

where u = 0 is the initial speed and t the time. Solving for t,

$t=\frac{v}{a}=\frac{0.5}{3.92}=0.128 s$

5 0

4 years ago

A 1-kg discus is thrown with a vertical velocity of 19 m/s at an angle of 35 degrees from a height of 1.94 m. Do not factor in a

Travka [436]

Answer:

Vertical velocity $V_y = 10.89 \ m/s$

Horizontal velocity $V_x = 15.57 \ m/s$

Explanation:

If a 1 -kg is thrown vertically with a velocity of 19 m/s at an angle of 35 °C from a height 1.94 m

The vertical and horizontal component can be resolved as:

$V_y = Vsin \theta \\ \\ \\\\V_x = Vcos \theta$

For Vertical component :

$V_y = V sin \theta \\\\V_y = 19 * sin 35 \\ \\ V_y = 19* 0..5736 \\ \\ V_y = 10.89 \ m/s$

For horizontal velocity

$V_x = V cos \theta \\\\V_x = 19 * cos 35 \\ \\ V_x = 19* 0.8192 \\ \\ V_x = 15.57 \ m/s$

4 0

3 years ago