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Black_prince [1.1K]
3 years ago
13

Consider the rocket in the image. Which unbalanced force accounts for the direction of the net force of the rocket ?

Physics
2 answers:
olya-2409 [2.1K]3 years ago
6 0

Answer:

Gravity

Explanation:

Gravity is bringing down the rocket and is not air resistance

Wittaler [7]3 years ago
3 0

Answer:

unbalenced force

Explanation:

make Brainliest  plz

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HELP PLEASE ASAP!!!<br> I CAN'T FAIL PHYSICS<br> xoxo thank you!
Allushta [10]
They are malleable and lustrous, and can conduct both electricity and thermal heat
5 0
3 years ago
Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
stepladder [879]

Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

Where:

F is the module of the force exerted between both bodies

G is the universal gravitation constant.

m_{1} and m_{2} are the masses of both bodies.

r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

5 0
3 years ago
A 50-g chunk of 80 degrees C iron is dropped into a cavity in a very large block of ice at 0 degrees C. Show that 5.5 g of ice w
Alenkasestr [34]

Answer:

5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

Explanation:

The concept to solve this problem is given by Energy Transferred, the equation is given by,

Q = mc\Delta T

Where,

Q= Energy transferred

m = mass of water

c = specific heat capacity

\Delta T = Temperature change (K or °C)

Replacing the values where mass is 50g and temperature is 80°C to 0°C we have,

Q = mc\Delta T

Q = 50*0.11*(80-0)

Q = 440cal

Then we can calculate the heat absorbed by m grams of ice at 0°C, then

Q_2 = mL = 80*m

How Q_1=Q_2, so

80m=440

m=\frac{440}{80}

m = 5.5g

Then 5.5g of ice melts when a 50g chunk of iron at 80°C is dropped into a cavity

7 0
4 years ago
Who were we in the space/arms race with?<br> In the movie *Hidden figures*
daser333 [38]

Explanation:

The U.S. launched its first man into space in May 1961.

4 0
3 years ago
A machine is rolling a metal cylinder under pressure. The radius of the cylinder is decreasing at a constant rate of 0.05 inches
Luden [163]
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
4 0
4 years ago
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