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3 years ago

In the parallelogram shown, AE = t + 2, CE = 3t − 14, and DE = 2t + 8.

Physics

2 answers:

solniwko [45]3 years ago

7 0

The answer would be 48

S_A_V [24]3 years ago

4 0

The answer is C 48 units

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NEED HELP TODAY

Free_Kalibri [48]

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

6 0

3 years ago

Read 2 more answers

A 25kg chair initially at rest on a horizontal floor requires a 165 N horizontal force to set it in motion. Once the chair is in

SCORPION-xisa [38]

Answer:

$\mu_k=0.51$

Explanation:

Given that

Mass , m = 25 kg

We know that when body is in rest condition then static friction force act on the body and when body is in motion the kinetic friction force act on the body .That is why these two forces are given as follows

Static friction force ,fs= 165 N

Kinetic friction force ,fk = 127 N

If the body is moving with constant velocity ,it means that acceleration of that body is zero and all the forces are balanced.

Lets take coefficient of kinetic friction = μk

The kinetic friction is given as follows

fk = μk m g

Now by putting the values

127 = μk x 25 x 9.81

$\mu_k=\dfrac{127}{25\times 9.81}$

$\mu_k=0.51$

Therefore the value of coefficient of kinetic friction will be 0.51

4 0

3 years ago

BRAINLEST FOR CORRECT ANSWER

xxMikexx [17]

Answer:

3kg sledgehammer swung at 1.5 m/s

Explanation:

Small Sledgehammer:

Mass:3.0

Velocity:1.5

MASS×VELOCITY=MOMENTUM

3.0×1.5= 4.5 (momentum)

Large Sledgehammer:

Mass:4.0

Velocity:0.9

4.0×0.9=3.6 (momentum)

higher momentum is the smaller Sledgehammer.

3 0

3 years ago

A 4.5 g coin sliding to the right at 23.8 cm/s makes an elastic head-on collision with a 13.5 g coin that is initially at rest.

Airida [17]

Answer:

a) $v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$ , b) $\Delta K = 9.559\times 10^{-5}\,J$

Explanation:

a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:

$(4.5\times 10^{-3}\,kg)\cdot (23.8\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot (0\,\frac{m}{s} ) = (4.5\times 10^{-3}\,kg)\cdot (-11.9\times 10^{-2}\,\frac{m}{s} )+(13.5\times 10^{-3}\,kg})\cdot v$

The final velocity is:

$v = 11.9\times 10^{-2}\,\frac{m}{s} \,(11.9\,\frac{cm}{s} )$

b) The change in the kinetic energy of the 13.5 g coin is:

$\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]$

$\Delta K = 9.559\times 10^{-5}\,J$

4 0

3 years ago

Read 2 more answers

What is the gravitational relationship between two objects?

Inessa [10]

I think the right answer would be objects pull because gravitational pull is when an object with more mass than an other object would pull the small mass object

5 0

3 years ago