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4 years ago

Please help. I need help with the questions on this page.

Physics

1 answer:

DaniilM [7]4 years ago

4 0

Try looking up facts about physics or referring back to your notes in class. :)
6025loveteal

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In February 2004, scientists at Purdue University used a highly sensitive technique to measure the mass of a vaccinia virus (the

Leno4ka [110]

2 hgkednexjurzvjoovcfcvvbvci

4 0

3 years ago

Two blocks A and B have a weight of 11 lb and 5 lb , respectively. They are resting on the incline for which the coefficients of

Alchen [17]

Answer:

$\theta=10.20^{\circ}$

$\Delta l=0.10 ft$

Explanation:

First of all, we analyze the system of blocks before starting to move.

$\Sum F_{x}=P_{A}sin(\theta)+P_{B}sin(\theta)-F_{fA}-F_{fB}=0$

$\Sum F_{x}=11sin(\theta)+5sin(\theta)-0.16N_{A}-0.23N_{B}=0$

$11sin(\theta)+5sin(\theta)-0.16P_{A}cos(\theta)-0.23P_{B}cos(\theta)=0$

$11sin(\theta)+5sin(\theta)-0.16*11cos(\theta)-0.23*5cos(\theta)=0$

$16sin(\theta)-2.91cos(\theta)=0$

$tan(\theta)=0.18$

$\theta=arctan(0.18)$

$\theta=10.20^{\circ}$

Hence, the incline angle θ for which both blocks begin to slide is 10.20°.

Now, if we do a free body diagram of block A we have that after the block moves, the spring force must be taken into account.

$P_{A}sin(\theta)-F_{fA}-F_{spring}=0$

Where:

$F_{spring} = k\Delta l=2.1\Delta l$

$P_{A}sin(\theta)-0.16*11cos(\theta)-2.1\Delta l=0$

$\Delta l=\frac{11sin(\theta)-0.16*11cos(\theta)}{2.1}$

$\Delta l=0.10 ft$

Therefore, the required stretch or compression in the connecting spring is 0.10 ft.

I hope it helps you!

4 0

4 years ago

A boy flies a kite with the string at a 30∘ angle to the horizontal. The tension in the string is 4.5 N. Part A Part complete Ho

zimovet [89]

Answer:

Work done is zero

Explanation:

given data

Angle of kite with horizontal = 30 degree

tension in the string = 4.5 N

WE KNOW THAT

Work = force * distance

horizontal force = $Tcos\theta = 4.5*cos30 = 3.89 N$

DISTANCE = 0 as boy stands still. therefore

work done = 3.89 *0 = 0

3 0

3 years ago

Someone talk o me I’m bored

stich3 [128]

Answer:

no plz its time to sleep byee

7 0

3 years ago

Read 2 more answers

Two plastic bowling balls, 1 and 2, are rubbed with cloth until they each carry a uniformly distributed charge of magnitude 0.50

galben [10]

Answer:

The right solution is " $4.5\times 10^{-10} \ Cm$ ".

Explanation:

Given that,

q = 0.50 nC

d = 900 mm

As we know,

⇒ $P=qd$

By putting the values, we get

⇒ $=0.50\times 900$

⇒ $=(0.50\times 10^{-9})\times 0.9$

⇒ $=4.5\times 10^{-10} \ Cm$

5 0

3 years ago

Read 2 more answers