All categories

2 years ago

Aye yo help me out plz

Physics

2 answers:

stiks02 [169]2 years ago

7 0

Answer:

B, Recoil after firing.

Explanation:

Lilit [14]2 years ago

3 0

Answer:

Explanation:

You might be interested in

A 9.0-kg bowling ball on a horizontal, frictionless surface experiences a net force of 6.0 n. what will be its acceleration?

Vladimir [108]

This question involves the concepts of Newton's Second Law of Motion.

The acceleration of the bowling ball will be "0.67 m/s²".

<h3>Newton's Second Law of Motion</h3>

According to Newton's Second Law of Motion, when an unbalanced force is applied on an object, it produces an acceleration in it, in the direction of the applied force. This acceleration is directly proportional to the force applied and inversely proportional to the mass of the object. Mathematically,

$F=ma\\\\a=\frac{F}{m}$

where,

a = acceleration = ?
F = Magnitude of the applied force = 6 N
m = Mass of the ball = 9 kg

Therefore,

$a=\frac{6\ N}{9\ kg}$

a = 0.67 m/s²

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525

#SPJ1

7 0

2 years ago

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$

6 0

3 years ago

A thermometer containing 0.10 g of mercury is cooled from 15 degrees celsius to 8.5 degrees celcius. How much energy left the me

loris [4]

To solve this exercise we will use the concept related to heat loss which is mathematically given as

$Q = mC_p \Delta T$

Where,

m = mass

$C_p$ = Specific Heat

$\Delta T =$ Change in temperature

Replacing with our values we have that

$m = 0.1g$

$C_p = 139J/Kg\cdot K \rightarrow$ Specific heat of mercury

$\Delta T = 8.5\°C-15\°C = -6.5\°C \Rightarrow -6.5K$

Replacing

$Q = (0.1*10^3)(138)(-6.5)\\Q = -0.09J$

Therefore the heat lost by mercury is 0.09J

5 0

3 years ago

According to the Big Bang theory, when the universe first formed, the temperature was extremely hot. As it began to cool, proton

alex41 [277]

hydrogen would be the first Element

5 0

3 years ago

An object will be stable if __________.

ryzh [129]

Answer:b

Explanation:

Object will be stable if its center of gravity lies over its base of support.

If the center of gravity is above the center of buoyancy then it provides positive righting moment and the object floats over the liquid and such object is said to possess positive meta-centric height.

6 0

3 years ago

Read 2 more answers