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3 years ago

10

A 0.500-kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to the object to change

the period to 2.00 s

1 answer:

Daniel [21]3 years ago

7 0

Answer:

389 kg

Explanation:

The computation of mass is shown below:-

$T = 2\pi \sqrt{\frac{m}{k} }$

Where K indicates spring constant

m indicates mass

For the new time period

$T^' = 2\pi \sqrt{\frac{m'}{k} }$

Now, we will take 2 ratios of the time period

$\frac{T}{T'} = \sqrt{\frac{m}{m'} }$

$\frac{1.50}{2.00} = \sqrt{\frac{0.500}{m'} }$

$0.5625 = \sqrt{\frac{0.500}{m'} }$

$m' = \frac{0.500}{0.5625}$

= 0.889 kg

Since mass to be sum that is

= 0.889 - 0.500

0.389 kg

or

= 389 kg

Therefore for computing the mass we simply applied the above formula.

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4. How long does it take a car traveling at 45 km/h to travel 100.0 m?<br> 4500m

Trava [24]

Answer:

8.0s

Explanation:

45 km/h ÷ 3.6 = 12.5 m/s

t.v = d/t

vt/v = d/v

t = d/v = 100.0m /12.5 m/s = 8.0s

Hope this helps!!

8 0

3 years ago

The Earth's escape speed (the speed you need to get away forever) is about 40,000 kilometers per hour. Escape speed depends on t

Anvisha [2.4K]

The Moon s escape speed will be smaller than Earth's.

What is escape speed:

The minimum speed that is required for an object to free itself from the gravitational force exerted by a massive object.

The formula of escape speed is

$v = \sqrt{\frac{2GM}{R} }$

where

v is escape velocity

G is universal gravitational constant

M is mass of the body to be escaped from

r is distance from the center of the mass

we can say that,

Escape speed depends on the gravity of the object trying to hold the spacecraft from escaping.

we know that,

The Moon's surface gravity is about 1/6th as powerful or about 1.6 meters per second per second.

since, v ∝ g

The Moon s escape speed will be smaller than Earth's.

Learn more about escape speed here:

<u>brainly.com/question/15318861</u>

#SPJ4

5 0

2 years ago

A lunch pail is accidentally kicked off a steel beam on a building under construction. Suppose the initial horizontal speed is 1

vichka [17]

1) 26.6 m

Along the horizontal direction, the lunch pail is moving with a uniform motion (constant speed), since there are no forces acting in this direction.

Therefore, the distance travelled horizontally after a time t is given by:

$d=v_x t$

where we know

$v_x = 1.50 m/s$ is the horizontal velocity

d = 3.50 m is the distance covered horizontally

Solving for t, we find the total time of the motion:

$t=\frac{d}{v_x}=\frac{3.50}{1.50}=2.33 s$

Now we know that the pail takes 2.33 s to fall to the ground. We can now consider the vertical motion of the pail, which is a free fall motion, so the vertical displacement is given by the equation

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Substutting t = 2.33 s, we find how fat the pail has fallen:

$s=\frac{1}{2}(9.8)(2.33)^2=26.6 m$

2) 10.7 m

In this case, we know instead the vertical displacement:

$s=2.50\cdot 10^2 m = 250 m$

Therefore, we can use the same equation again

$s=ut+\frac{1}{2}at^2$

To find the total time of motion:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(250)}{9.8}}=7.14 s$

We know that along the horizontal direction, the velocity is constant:

$v_x = 1.50 m/s$

So, the horizontal distance covered in this time is

$d=v_x t = (1.50)(7.14)=10.7 m$

6 0

4 years ago

Explain how an intrusive igneous rock could become a metamorphic rock and then an extrusive igneous rock.

sleet_krkn [62]

1.igneous rocks are formed when magma or lava cools and hardens
2. the cooling rate of the rock

5 0

3 years ago

docker41 [41]

<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

mass, $\sf{m_c=40\ g.}$

specific heat capacity, $\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_c=200^{\circ}C.}$

Then the heat of copper :

$\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}$

$\sf{\dashrightarrow Q_c =16(T-200)\ J}$

For copper calorimeter :

mass, $\sf{m_{cc} =60\ g.}$

specific heat capacity, $\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_{cc} =25^{\circ}C.}$

Then the heat of copper calorimeter :

$\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}$

$\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}$

For water :

mass, $\sf{m_w=50\ g. }$

specific heat capacity, $\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_w=25^{\circ}C.}$

Then heat of water :

$\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}$

$\sf{\dashrightarrow Q_w=210(T-25)\ J}$

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

$\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}$

$\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}$

$\sf{\dashrightarrow 250T- 9050=0}$

$\sf{\dashrightarrow T=36.2^{\circ}C}$

$\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}$

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from $\sf{^{\circ}C}$ to K or K to $\sf{^{\circ}C}$ .]

7 0

3 years ago