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3 years ago

Two forces, F1 and F2, are applied to a block on a frictionless, horizontal surface

Physics

2 answers:

wel3 years ago

7 0

Answer:5kg

Explanation:I got it right on my test

Komok [63]3 years ago

5 0

Answer:

F, = 12N. F, = 2 N. Block. 4) a 20.0-kg mass moving at 1.00 m/s.

Explanation:

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An electron has a constant acceleration of +3.3 m/s2. at a certain instant its velocity is +8.6 m/s. what is its velocity (a) 2.

Citrus2011 [14]

Hello!

The velocity function is:

$\mathsf{V = V_0 + at}\\ \\ \mathsf{V = 8.6 +3.3t}$

a) 2.3 s earlier, we can make t = -2.3 s, in relation to our initial time:

$\mathsf{V = 8.6 + 3.3\cdot(-2.3)}\\ \\ \mathsf{V = 8.8-7.59}\\ \\ \boxed{\mathsf{V = 1.21 \ m/s}}$

b) Now we can make t = 2.3 s:

$\mathsf{V = 8.6 + 3.3\cdot2.3}\\ \\ \mathsf{V = 8.6 + 7.59}\\ \\ \boxed{\mathsf{V = 15.59 \ m/s}}$

7 0

3 years ago

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction betwe

Gre4nikov [31]

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

5 0

3 years ago

find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds

Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, $u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}$

Final speed of the train, $v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}$

Displacement of the train, $S=234\textrm{ km}=234\times 1000=234000\textrm{ m}$

Using Newton's equation of motion,

$v - u = at\\a=\frac{v-u}{t}$

Now, using Newton's equation of motion for displacement,

$v^{2}-u^{2}=2aS$

Now, plug in the value of $a=\frac{v-u}{t}$ in the above equation. This gives,

$v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}$

Now, plug in 234000 m for $S$ , 25 m/s for $v$ and 20 m/s for $u$ . Solve for $t$ .

$t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}$

Therefore, the time taken by the train is 10400 s.

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4 years ago

A copper block rests 30.0 cm from the center of a steel turntable. The coefficient of static friction between the block and the

PIT_PIT [208]

Answer:

refer to the above attachment

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1 year ago

When practicing an oral presentation what can you do to prepare for the presentation

Igoryamba

When practicing an oral presentation, you can prepare by writing a draft and practice reading aloud what you are going to say before your oral presentation.

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3 years ago