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vlada-n [284]
4 years ago
9

What's 7y^2 - y^2? HELP!

Physics
2 answers:
trapecia [35]4 years ago
8 0
Hey there...
Let's solve this problem by steps.

Method 1 -
Step 1 --> Taking GCF
y^2 ( 7 - 1 )

Step 2 ---> Combining like terms
y^2 ( 6)

Or

6y^2  is your final answer.

Method 2 -
We can straight way subtract y^2 from 7y^2 which is
7y^2 - y^2
=\ \textgreater \  6y^2 

Hope it helped you..  :)
34kurt4 years ago
6 0
<span>7y^2 - y^2 = 6y^2 i think!</span>
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A ball is hit 1 meter above the ground with an initial speed of 40 m/s. If the ball is hit at an angle of 30 above the horizonta
Vika [28.1K]

Answer:

t_t=4.131\ s

Explanation:

Given:

height above the horizontal form where the ball is hit, y=1\ m

angle of projectile above the horizontal, \theta=30^{\circ}

initial speed of the projectile, u=40\ m.s^{-1}

<u>Firstly we find the </u><u>vertical component of the initial velocity</u><u>:</u>

u_y=u.\sin\theta

u_y=40\times \sin30^{\circ}

u_y=20\ m.s^{-1}

During the course of ascend in height of the ball when it reaches the maximum height then its vertical component of the velocity becomes zero.

So final vertical velocity during the course of ascend:

  • v_y=0\ m.s^{-1}

Using eq. of motion:

v_y^2=u_y^2-2g.h (-ve sign means that the direction of velocity is opposite to the direction of acceleration)

0^2=20^2-2\times 9.8\times h

h=20.4082\ m (from the height where it is thrown)

<u>Now we find the time taken to ascend to this height:</u>

v_y=u_y-g.t

0=20-9.8t

t=2.041\ s

<u>Time taken to descent the total height:</u>

  • we've total height, h'=h+y =20.4082+1

h'=u_y'.t'+\frac{1}{2} g.t'^2

  • during the course of descend its initial vertical velocity is zero because it is at the top height, so u_y'=0\ m.s^{-1}

21.4082=0+4.9t'^2

t'=2.09\ s

<u>Now the total time taken by the ball to hit the ground:</u>

t_t=t'+t

t_t=2.09+2.041

t_t=4.131\ s

3 0
3 years ago
A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.40 kg box that is sitting on the horizontal,
Fynjy0 [20]

Answer:v=7.24 m/s

Explanation:

Given

F(x)=18-0.530 x

mass m=8.40 kg

box is at rest at x=0

Distance traveled=16 m

mv\frac{\mathrm{d} v}{\mathrm{d} x}=F=\left ( 18-0530x\right )dx

\int_{0}^{v}vdv=\int_{0}^{16}\left ( 18-0.530x\right )dx

\frac{mv^2}{2}=\left ( 18x-0530\frac{x^2}{2}\right )_0^{16}

v^2=\frac{288-67.841\times 2}{8.40}

v=7.24 m/s

6 0
3 years ago
A dog sledding team is comprised of a 80.0 kg musher (the person driving the sled), a 21.0 kg sled and four dogs. assume each do
Xelga [282]
You would divide or multiply by 9.8
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4 years ago
A proton moves with a speed of 0. 985c. (a) calculate its rest energy
mojhsa [17]

The rest energy of the proton will be 938 MeV. The rest energy is related to the rest mass of the proton.

<h3>What is rest energy?</h3>

The rest energy is the rest mass times the square of the speed of light of a particle at rest in an inertial frame of reference.

The rest energy is found as;

\rm E_R= mC^2\\\\  E_R=(1.67 \times 10^{-27})\times 3 \times 10^8)^2\\\\ E_R= 9.375 \times 10^8 / eV

Convert into the mega electron volt;

\rm E_R= 938 \  MeV

Hence, the rest energy of the proton will be 938 MeV

To learn more about the rest energy, refer to the link;

brainly.com/question/15047418

#SPJ4

4 0
2 years ago
An airplane takes off from Boston for
Ede4ka [16]

Answer:B

Explanation:

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t=2h ---------------------   time

v=?

------------------------

V=S/t

V=980km/2h

V=490km/h

7 0
4 years ago
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