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2 years ago

Which parameter of a projectile depends on the horizontal as well as the vertical component of velocity of projection?

Physics

2 answers:

Maurinko [17]2 years ago

7 0

Answer:

It RANGE

Explanation:

I just did it and that's the right answer. Angle of launch creates the horizontal and vertical velocity components, it is not dependent on them!!!!

jekas [21]2 years ago

6 0

Just about anything you could ask about a projectile AFTER it's launched
depends on both components of the launch velocity.

Here are some that I can think of:

-- angle of launch
-- magnitude of launch velocity
-- location at any time after launch
-- magnitude of velocity at any time after launch
-- direction of velocity at any time after launch
-- distance of the landing point from the launch point

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11. A vector M is 15.0 cm long and makes an angle of 20° CCW from x axis and another vector N is 8.0 cm long and makes an angle

Alja [10]

Answer:

The magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

Explanation:

To find the resultant vector, you first calculate x and y components of the two vectors M and N. The components of the vectors are calculated by using cos and sin function.

For M vector you obtain:

$M=M_x\hat{i}+M_y\hat{j}\\\\M=15.0cm\ cos(20\°)\hat{i}+15.0cm\ sin(20\°)\hat{j}\\\\M=14.09cm\ \hat{i}+5.13\ \hat{j}$

For N vector:

$N=N_x\hat{i}+N_y\hat{j}\\\\N=8.0cm\ cos(40\°)\hat{i}+8.0cm\ sin(40\°)\hat{j}\\\\N=6.12cm\ \hat{i}+5.142\ \hat{j}$

The resultant vector is the sum of the components of M and N:

$F=(M_x+N_x)\hat{i}+(M_y+N_y)\hat{j}\\\\F=(14.09+6.12)cm\ \hat{i}+(5.13+5.142)cm\ \hat{j}\\\\F=20.21cm\ \hat{i}+10.27cm\ \hat{j}$

The magnitude of the resultant vector is:

$|F|=\sqrt{(20.21)^2+(10.27)^2}cm=22.66cm$

And the direction of the vector is:

$\theta=tan^{-1}(\frac{10.27}{20.21})=29.93\°$

hence, the magnitude of the resultant vector is 22.66 cm and it has a direction of 29.33°

4 0

3 years ago

Two violinists are trying to tune their instruments in an orchestra. One is producing the desired frequency of 440.0 hz. The oth

Katyanochek1 [597]

Answer:

Percentage change in tension is 3.8%

Explanation:

We have given initially frequency $f_1$ = 440 Hz

Let tension in the string at this frequency is $T_1$

Now second frequency is $f_2=448.4Hz$

Frequency in string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

From the relation we can say that

$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$

$\frac{440}{448.4}=\sqrt{\frac{T_1}{T_2}}$

${\frac{T_2}{T_1}}=1.038$

Percentage change in tension is equal to

$=\frac{T_2-T_1}{T_1}=\frac{T_2}{T_1}-1=(1.038-1)\times 100=3.8$ %

So percentage change in tension is 3.8%

4 0

3 years ago

The fragment of an asteroid or any interplanetary material is known as a: A) limestone dignitary satellite. B) moon. C) shower m

ElenaW [278]

The fragment of an asteroid or any interplanetary material is known as a a : D. Meteroid

Human came in contact with this material mostly because it penetrate the atmosphere and fall within the earth surface

hope this helps

6 0

3 years ago

What is the formula for work?

babunello [35]

W = F • d • cos(theta)

6 0

3 years ago

Two boys are sliding toward each other on a frictionless, ice-covered parking lot. Jacob, mass 45 kg, is gliding to the right at

Maurinko [17]

Answer:

$v=0.3381\ m.s^{-1}$ in the direction of motion of Jacob