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2 years ago

A 20 kg weight is suspended from a string. What is the tension on the string

2 answers:

7 0

The weight is in equilibrium, so the two forces acting on it (tension T pulling upward, and its own weight W pulling it downward) cancel each other. By Newton's second law,

∑ F = T - W = 0

so that

T = W

Calculate the object's weight:

W = (20 kg) (9.8 m/s²) = 196 N

So the tension is T = 196 N as well.

JulsSmile [24]2 years ago

3 0

assume that tension in string a is Ta and in string b is Tb

now taking the component of tensions

Tasin30∘=Tbsin30∘...(1)

it means Ta=Tb

Tacos30∘+Tbcos30∘=mg...(2)

⇒2Tacos30∘=20×10⇒Ta=Tb=3200

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Someone help please by providing work and answers please :)

cupoosta [38]

There are two ways to solve this. The longer way is to use those equations to calculate numbers for total distance.

The easier way is to find the area under the graph. That's right, AREA UNDER VELOCITY-TIME graph is the TOTAL DISTANCE travelled!

it's a shortcut.

Let's split up the area into a triangle and rectangle:

Triangle = 0.5(4-0)(10-0) = 20 m
Rectangle = (6-4)(10-0) = 20 m

Total distance = 40 m!

6 0

3 years ago

A newly discovered particle, the SPARTYON, has a mass 945 times that of an electron. If a SPARTYON at rest absorbs an anti-SPART

Gennadij [26K]

Answer:

$\nu =1166\times 10^{20}Hz$

Explanation:

We have given the rest mass of SPARTYON = 945 times of mass of electron

We know that mass of electron $=9.11\times 10^{-31}kg$

So mass of SPARTYON $=945\times 9.11\times 10^{-31}=8608.95\times 10^{-31}kg$

Speed of light $c=3\times 10^8m/sec$

According to Einstein equation energy is given by

$E=mc^2=8608.95\times 10^{-31}\times (3\times 10^{8})^2=77480.55\times 10^{-15}j$

Now according to planks's rule

Energy is given by

$E=h\nu$ , here h is plank's constant $h=6.6\times 10^{-34}$

So $77480.55\times 10^{-15}=6.6\times 10^{-34}\nu$

$\nu =1166\times 10^{20}Hz$

3 0

3 years ago

Why does it take double the applied force to move a mass double the size?

Ratling [72]

56-999999999999999999999-4 is the best for my mom

5 0

2 years ago

To navigate, a porpoise emits a sound wave that has a wavelength of 3.3 cm. The speed at which the wave travels in seawater is 1

dedylja [7]

Answer:

$2.2\times 10^{-5} s$

Explanation:

We are given that

The wavelength of sound wave= $\lambda=3.3 cm=3.3\times 10^{-2}m/s$

1 cm/s= $10^{-2}m/s$

Speed of sound wave,v= $1522 m/s$

We have to find the period of the wave.

We know that

Frequency= $\nu=\frac{v}{\lambda}$

Using the formula

Frequency = $\frac{1522}{3.3\times 10^{-2}}=4.6\times 10^{4}$ Hz

Time period= $\frac{1}{4.6\times 10^4}=0.22\times 10^{-4}\times \frac{10}{10^1}=2.2\times 10^{-4-1}=2.2\times 10^{-5}s$

Using identity: $\frac{a^x}{a^y}=a^{x-y}$

Hence, the time period of the wave= $2.2\times 10^{-5} s$

4 0

2 years ago

A pendulum at position A is released and swings through position B to position Con the other side.

zaharov [31]

Explanation:

Given the conditions A,B and C when the pendulum is released, at point A the initial velocity of the pendulum is zero(0), the potential energy stored is maximum(P.E= max),

the conditions can be summarized bellow

point A

initial velocity= 0

final velocity=0

P.E= Max

K.E= 0

point B

initial velocity= maximum

final velocity=maximum

P.E=K.E

point C

initial velocity= min

final velocity=min

P.E= 0

K.E= max

3 0

2 years ago