All categories

3 years ago

PLEASE HELPPPPPPPPPPP I WILL PICK BRAINLIEST ANSWER THANKS

Physics

1 answer:

Lemur [1.5K]3 years ago

7 0

Been a while since i've done a problem like this so i might be wrong in the "work" aspect

your system starts with 1200J

you add 700J of heat-
add 700 to 1200
1200+700

and your system does 400J of work
since work is being done by the system, the energy is leaving the system hence you subtract
1900-400

the result is then 1500

You might be interested in

Sunitha can type 1800 words in half an hour. What is her typing speed in words per minute?

Andre45 [30]

Answer:

60words/minute

Explanation:

If Sunitha can type 1800 words in half an hour, this can be expressed as;

1800 words = 30 minutes

To get her typing speed per minute, we will use the formula

Speed = Number of words/Time used

Typing speed = 1800/30

Typing speed = 60words/minute

Hence her typing speed in words per minute is 60words/minute

6 0

2 years ago

Which statement best describes frustration?

wlad13 [49]

C. when you can't achieve your goal due to events beyond your control

5 0

3 years ago

As an intern with an engineering firm, you are asked to measure the moment of inertia of a large wheel, for rotation about an ax

AysviL [449]

Answer:

$I=2.766\ kg.m^2$

Explanation:

We have:

diameter of the wheel, $d=0.88\ m$

weight of the wheel, $w_w=280\ N$

mass of hanging object to the wheel, $m_o=6.32\ kg$

speed of the hanging mass after the descend, $v_o=4\ m.s^{-1}$

height of descend, $h=2.5\ m$

(a)

moment of inertia of wheel about its central axis:

$I=\frac{1}{2} m.r^2$

$I=\frac{1}{2} \frac{w_w}{g}.r^2$

$I=\frac{1}{2} \times \frac{280}{9.8}\times 0.44^2$

$I=2.766\ kg.m^2$

3 0

3 years ago

A total Δν of 15 km/s is required to achieve an interplanetary mission. The proposed rocket has two stages. The first stage alon

AlexFokin [52]

Answer:

102000 kg

Explanation:

Given:

A total Δν = 15 km/s

first stage mass = 1000 tonnes

specific impulse of liquid rocket = 300 s

Mass flow rate of liquid fuel = 1500 kg/s

specific impulse of solid fuel = 250 s

Mass flow of solid fuel = 200 kg/s

First stage burn time = 1 minute = 1 × 60 seconds = 60 seconds

Now,

Mass flow of liquid fuel in 1 minute = Mass flow rate × Burn time

Mass flow of liquid fuel in 1 minute = 1500 × 60 = 90000 kg

Also,

Mass flow of solid fuel in 1 minute = Mass flow rate × Burn time

Mass flow of solid fuel in 1 minute = 200 × 60 = 12000 kg

Therefore,

The total jettisoned mass flow of the fuel in first stage

= 90000 kg + 12000 kg

= 102000 kg

3 0

3 years ago

Find the angle formed by two forces of 7N and 15N respectively if its result is worth 20N

nadezda [96]

First, you need to make certain assumptions before solving this question. Why? Because there are no information given about the direction of these forces. In such questions as above, ALWAYS make the following assumptions:

1) Take first force, say $F_{1}$ , and assume that it is pointing towards the x-direction.

Let us take the 7N force! By keeping the above assumption in our minds, the force vector would be like:
$F_{1}$ = $7i$ , where $i$ = Unit vector in the x-direction.

2) Take the second force, say $F_{2}$ , and assume that it is making an angle $\alpha$ with the first force $F_{1}$ .

Let us take the 15N force! By keeping the above assumption in our minds, the forces vector would be like:

$F_{2} = (15*cos \alpha)i + (15*sin \alpha )j$

Now from simple vector addition, we know that,
$F_{R} = F_{1} + F_{2}$ --- (A)

Where $F_{R}$ = Resultant vector.
NOTE: In equation (A), all forces are in vector notation. Assume that there is an arrow head on top of them.

Let us find $F_{1}+F_{2}$ first!
$F_{1}+F_{2}$ =   $7i+(15*cos \alpha)i + (15*sin \alpha )j$

=> $F_{1}+F_{2}$ =   $(7+15*cos \alpha)i + (15*sin \alpha )j$

Now the magnitude of $F_{1}+F_{2}$ is,
| $F_{1}+F_{2}$ | = $\sqrt{ (7+ 15*cos \alpha)^{2} + (15*sin \alpha )^{2}}$

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225*(cos \alpha)^{2} + 210*(cos \alpha)+ 255*(sin \alpha )^{2}}$

Since $(sin \alpha)^{2} + (cos \alpha)^{2} = 1$ , therefore,

=> | $F_{1}+F_{2}$ | = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Since  | $F_{1}+F_{2}$ | = | $F_{R}$ |, and the magnitude of the resultant force is 20N, therefore,

| $F_{R}$ | = | $F_{1}+F_{2}$ |
20 = $\sqrt{ 49 + 225 + 210*(cos \alpha)}$

Take square on both sides,
400 = $49 + 225 + 210*(cos \alpha)$
$(cos \alpha) = \frac{3}{5}$

$\alpha = 53.13^{o}$

Ans: Angle formed by the two forces, 7N and 15N, is: 53.13°

-israr

4 0

2 years ago