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2 years ago

Please, pretty please help me!

Physics

2 answers:

Andrews [41]2 years ago

8 0

Answer:

umm

Explanation:

1223565 578633 =334675

Harrizon [31]2 years ago

7 0

Hhsjjsjdbdvbr fineee

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A tomato of mass 0.18 kg is dropped from a tall bridge. If the tomato has a speed of 11 m/s just before it hits the ground, what

kondor19780726 [428]

The kinetic energy of the tomato is :

K.E = 1/2 mv^2

K.E = 1/2 x 0.18 kg x 11 m/S^2

K.E = 0.99

Hope this helps

7 0

3 years ago

Apply the general results obtained in the full analysis of motion under the influence of a constant force in Section 2.5 to answ

zvonat [6]

Answer:

y(i) = h

v(y.i) = 0

Explanation:

See attachment for elaboration

3 0

3 years ago

A skater extends her arms, holding a 2 kg mass in each hand. She is rotating about a vertical axis at a given rate. She brings h

Usimov [2.4K]

Explanation:

It is known that relation between torque and angular acceleration is as follows.

$\tau = I \times \alpha$

and, I = $\sum mr^{2}$

So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

= 4 $kg m^{2}$

$\tau_{1} = 4 kg m^{2} \times \alpha_{1}$

$\tau_{2} = I_{2} \alpha_{2}$

So, $I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}$

= 1 $kg m^{2}$

as $\tau_{2} = I_{2} \alpha_{2}$

= $1 kg m^{2} \times \alpha_{2}$

Hence, $\tau_{1} = \tau_{2}$

$4 \alpha_{1} = \alpha_{2}$

$\alpha_{1} = \frac{1}{4} \alpha_{2}$

Thus, we can conclude that the new rotation is $\frac{1}{4}$ times that of the first rotation rate.

8 0

3 years ago

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

so:

$\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m$

3 0

3 years ago

What is the period if the block’s mass is doubled? note that you do not know the value of either m or k , so do not assume any p

Oksanka [162]

The period of the block's mass is changed by a factor of √2 when the mass of the block was doubled.

The time period T of the block with mass M attached to a spring of spring constant K is given by,

T = 2π(√M/K).

Let us say that, when we increased the mass to 2M, the time periods of the block became T', the spring constant is not changed, so, we can write,

T' = 2π(√2M/K)

Putting T = 2π(√M/K) above,

T' =√2T

So, here we can see, if the mass is doubled from it's initial value. The time period of the mass will be changed by a factor of √2.

To know more about time period of mass, visit,

brainly.com/question/20629494

#SPJ4

5 0

1 year ago