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kolbaska11 [484]

2 years ago

10

Please Please Please Can Help Me On This Question!!!!! I Give Thanks!!!! Please Do 1-4!!!!

2 answers:

ololo11 [35]2 years ago

5 0

1.add the amount of the diagram which is M+Y then dived the answer you get.

yuradex [85]2 years ago

4 0

Answer:

add the amount of the diagram which is M+Y then dived the answer

Explanation:

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Who’s good at algebra?

lutik1710 [3]

Answer:

I'm alright................

5 0

3 years ago

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Suppose a soccer ball is kicked from the ground at an angle 20.0º above the horizontal at 8.00 m/s. The y-velocity is determined

julia-pushkina [17]

initial speed of the ball by which it is kicked is 8 m/s

now the angle at which it is kicked is 20 degree

here we will have

$v_y = v sin20$

$v_y = 8 sin20 = 2.74 m/s$

now we will have

$\Delta y = v_y t + \frac{1}{2}at^2$

so if the ball again land on the ground at same level then we have

$\Delta y = 0$

$0 = 2.74 t - \frac{1}{2}(9.8) t^2$

$0 = 2.74 - 4.9 t$

$t = 0.56 s$

so total time will be 0.56 s

8 0

3 years ago

A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring

azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 $\frac{N}{m}$

Angle $\theta$ = 34°

From the free body diagram

Force acting on the box = mg sin $\theta$

⇒ F = 3 × 9.81 × $\sin34$

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

$F_{sp}$ = k x

$F_{sp}$ = 120 $x$ -------- (2)

The net force acting on the body is given by

$F_{net} = ma$

Since acceleration of the box is zero so

$F_{net} = 0$

$F - F_{sp} = 0$

$F = F_{sp}$

Put the values from equation (1) & (2) we get

16.45 = 120 $x$

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0

3 years ago

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

What is the awnser to that question

Olegator [25]

The answers To your question is c

8 0

3 years ago

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