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tangare [24]
3 years ago
12

What are two things needed for acceleration

Physics
1 answer:
Liula [17]3 years ago
5 0

Answer:

Velocity and Time

Explanation:

I googled it-

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A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.
Bogdan [553]

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s

Momentum = 310*0.022

Momentum = 6.82 Ns.

If the bullet gets embedded they will both have common velocity v

6.82 = (0.022+1.40)v

6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

Where tita is 90°

H = (4.796² * sin²(90))/(2*9.81)

H =( 23.001616*1)/19.62

H = 1.172 metres

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4 years ago
A spring tide can occur at what day and month?
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Spring tides occur twice each lunar month all year long (they don't depend on any particular season. When there is a new or full moon the average tidal ranges are larger.
7 0
3 years ago
Why are bridges declared unsafe after a long use​
lina2011 [118]

After it's long use, the bridge continuously undergoes through alternating strains. in result, the elastic strength of the bridge gets reduced. while the process goes on the bridge makes elastic fatigue and a permanent change happens which makes it collapse which is the reason why bridges are declared unsafe after a long use​ :)

8 0
3 years ago
As computer information technology becomes faster, how does the ability of scientists to collect and organize data change?
irina1246 [14]
Because computers and technology develop and become faster things are going to<span> become substantially easier.</span>
7 0
4 years ago
If the ball leaves the projectile launcher at a speed of 2.2 m/s at an angle of 30ᴼ, and the projectile launcher is on a table a
IgorC [24]

Answer: 1.12 m

Explanation:

This situation is related to parabolic motion, hence we can use the following equations:

y=y_{o}+V_{o}sin \theta t-\frac{g}{2}t^{2} (1)

x=V_{o} cos \theta t (2)

Where:

y=0 m is the ball final height (when it hits the ground)

y_{o}=1.1 m is the ball initial height

V_{o}=2.2 m/s is the initial velocity

\theta=30\° is the angle at which the ball was launched

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the horizontal distance the ball travels

Rewriting (1) with the given values:

0 m=1.1 m+(2.2 m/s)(cos 30\°)t-\frac{9.8 m/s^{2}}{2}t^{2} (3)

Multiplying all the eqquation by -1 and rearranging:

4.9 m/s^{2} t^{2}-1.1 m/s t-1.1 m=0 (4)

So, since we have a quadratic equation here (in the form of0=at^{2}+bt+c,  we will use the quadratic formula to find  t:  

t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}   (5)

Where a=4.9, b=-1.1, c=-1.1  

Substituting the known values and choosing the positive result of the equation, we have:  

t=\frac{-(-1.1)\pm\sqrt{(-1.1)^{2}-4(4.9)(-1.1)}}{2(4.9)}  

t=0.59 s (6)

Now, substituting (6) in (2):

x=(2.2 m/s)(cos 30\°)(0.59 s) (7)

x=1.12 m (8) This is the horizontal distance at which the ball hits the ground.

3 0
4 years ago
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