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2 years ago

15

suppose that you want to build a machine to perform simple tasks. why must you understand all about forces to complete your miss

ion?

1 answer:

mixas84 [53]2 years ago

4 0

Answer:

CAUSE YPU WANT TO

Explanation:

You might be interested in

Question 6 (10 points)

horsena [70]

Answer:

a= g = - 9.81 m/s2.

The following equations will be helpful:

a = (vf - vo)/t d = vot + 1/2 at2 vf2 = vo2 + 2ad

When you substitute the specific acceleration due to gravity (g), the equations are as follows:

g = (vf - vo)/t d = vot + 1/2 gt2 vf2 = vo2 + 2gd

If the object is dropped from rest, the initial velocity ("vi") is zero. This further simplifies the equations to these:

g = vf /t d = 1/2 gt2 vf2 = 2gd

The sign convention that we will use for direction is this: "down" is the negative direction. If you are given a velocity such as -5.0 m/s, we will assume that the direction of the velocity vector is down. Also if you are told that an object falls with a velocity of 5.0 m/s, you would substitute -5.0 m/s in your equations. The sign convention would also apply to the acceleration due to gravity as shown above. The direction of the acceleration vector is down (-9.81 m/s2) because the gravitational force causing the acceleration is directed downward.

hope this info helps you out!

7 0

2 years ago

Paleontologists are experts on the rock cycle true or false

kow [346]

Answer:

I cannot do the laws of physics cuz I hate science but I'm just dance

Explanation:

<u>it's</u><u> it's true</u>

8 0

2 years ago

I was reading an old thermodynamics textbook and came across this equation describing change in internal energy. What does the (

RoseWind [281]

Explanation:

I remember that notation! The expression

$dQ = dU = (\dfrac{\partial U}{\partial T})_{V} dT+ (\dfrac{\partial U}{\partial V})_{T}dV$

is the 1st law of thermodynamics and it refers to the heat supplied to the system dQ which is also a change in its internal energy dU. The first term is the <u>partial</u> derivative of the internal energy U with respect to temperature T while the volume V is kept constant, as denoted by the subscript V. The 2nd term is similar but this time, temperature is kept constant while its volume partial derivative is being taken.

Ah, memories!

8 0

2 years ago

A boy can swim 3.0 meter a second in still water while trying to swim directly across a river from west to east, he is pulled by

lana66690 [7]

Answer:

Angle: $48.19^o$

Explanation:

<u>Two-Dimension Motion</u>

When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.

Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.

To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is

$v_b\ cos\alpha$

where $v_b$ is the speed of the boy in still water and $\alpha$ is the angle respect to the shoreline. If the river flows at speed $v_s$ , we now set

$v_b\ cos\alpha=v_s$

$\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}$

$\alpha=48.19^o$

8 0

2 years ago

A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s

Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g= $4.67\times 10^{-3} kg$

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1, $m_1=1177g=1.177kg$

Mass of block 2, $m_2=1626 g=1.626 kg$

Velocity of block 1, $v_1=0.681m/s$

(a)

Let velocity of the second block after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

$mv=m_1v_1+(m+m_2)v_2$

$4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2$

$1.66719=0.801537+1.63067v_2$

$1.66719-0.801537=1.63067v_2$

$0.865653=1.63067v_2$

$v_2=\frac{0.865653}{1.63067}$

$v_2=0.531m/s$

Hence, the velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

$K_i=\frac{1}{2}mv^2$

$k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)$

$k_i=297.59 J$

Final kinetic energy after collision

$K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2$

$K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2$

$K_f=0.5028 J$

Now, he ratio of the total kinetic energy after the collision to that before the collision

= $\frac{k_f}{k_i}=\frac{0.5028}{297.59}$

=0.00169

5 0

2 years ago