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2 years ago

Give three examples of objects in equilibrium found in classroom?

Physics

1 answer:

Charra [1.4K]2 years ago

7 0

Answer:

Book. Bottle. table are some examples of objects in equilibrium found in classroom

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A 265 g mass attached to a horizontal spring oscillates at a frequency of 3.40 Hz . At t =0s, the mass is at x= 6.20 cm and has

lara [203]

Answer:

The phase constant is 7.25 degree

Explanation:

given data

mass = 265 g

frequency = 3.40 Hz

time t = 0 s

x = 6.20 cm

vx = - 35.0 cm/s

solution

as phase constant is express as

y = A cosФ ..............1

here A is amplitude that is = $\sqrt{(\frac{v_x}{\omega })^2+y^2 }$ = $\sqrt{(\frac{35}{2\times \pi \times y})^2+6.2^2 }$ = 6.25 cm

put value in equation 1

6.20 = 6.25 cosФ

cosФ = 0.992

Ф = 7.25 degree

so the phase constant is 7.25 degree

5 0

3 years ago

A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

$p = \frac{fd}{t} \\ where \: \\p = power \\ f = force \\ d = distance \\ and \: t = time \\ \\ p = \frac{3800 \times 50}{10} \\ p = \frac{190000}{10} \\ p = 19000w$

7 0

3 years ago

An eastbound car travels a displacement of 80.0 Km in 45 minutes. What is the velocity

alexgriva [62]

Explanation:

d= 80km = 8000m

t = 45 min = 45/60 h

= 0.75 h

V= ?

we know that,

V = d /t

or,V= 80 km / 0.75 h

or, V= 106.67 km/hr

or,V= 106.67×1000m / 3600 s

2. or, V= 29.63 m/s

4 0

2 years ago

The instantaneous speed of a particle moving along one straight line is v(t) = ate−6t, where the speed v is measured in meters p

beks73 [17]

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

$v(t)=ate^{-6t}$ (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

$\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]$ (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

$a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}$

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

$v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a$

hence, the maximum speed is v_max = ((1/6)e^-1)a

5 0

3 years ago

A drop

exis [7]

Answer:

$27.5\ m$

Explanation:

As we know that volume of cylinder is

$v=\pi r^{2} *h$

Where v=volume , h= height or thickness and r= radius

Here,

$v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5$

Putting these values in the previous equation , we get

$10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m$

Therefore thickness is 27.5 m

7 0

3 years ago

Give three examples of objects in equilibrium found in classroom?​

Give three examples of objects in equilibrium found in classroom?