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aliina [53]
3 years ago
5

I drop a meter stick , and catch it as it falls. If it fell exactly 22 cm, how much did it take me to catch the meter stick (wha

t was my reaction time)?
Physics
1 answer:
statuscvo [17]3 years ago
3 0
D=Vot+1/2at^2

In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2

acceleration=acceleration due to gravity=-9.8m/s^2

It falls - 22cm or -0.22m

We have - 0.22=1/2(-9.8)t^2

t^2=(-0.44)/(-9.8)

t=sqrt[0.44/9.8]
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You are trying to hear your friend give directions to new store in town. But from your distance (1 point) of 15 m you only hear
Marat540 [252]

Answer:

option D

Explanation:

given,

Intensity of sound = 20 dB

distance = 15 m

intensity of sound is increased to = 50 dB

distance between the sound level = ?

Using relation

L_2 = L_1 - |20(log \dfrac{r_2}{r_1})|

L₁ = 20 dB        L₂ = 50 dB         r₁ = 15 m      r₂ = ?

log (\dfrac{r_2}{r_1}) = \dfrac{L_1 -L_2}{20}

\dfrac{r_2}{r_1}= 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =r_1 10^{\dfrac{|L_1 -L_2|}{20}}

r_2 =15 \times 10^{\dfrac{|20-50|}{20}}

r_2 =15 \times 10^{-1.5}

r₂ = 0.47 m

r₂ = 47 cm

hence, the correct answer is option D

7 0
3 years ago
One type of potential energy is gravitational potential energy. What is necessary for an object to have gravitational potential
galben [10]

Answer:

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. Since the force required to lift it is equal to its weight, it follows that the gravitational potential energy is equal to its weight times the height to which it is lifted.

5 0
3 years ago
Read 2 more answers
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which
julia-pushkina [17]

Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = 11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = 390\ W/m.^{\circ}C

Conduction of heat from the rod per second is given by:

q = \frac{KA\Delta T}{L}

where

\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C = temperature difference between the two ends of the rod.

Thus

q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s

Now,

To calculate the mass, M of the ice melted per sec:

M = \frac{q}{L_{w}}

where

L_{w} = Latent heat of fusion of water = 333 kJ/kg

M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g

5 0
3 years ago
A car travels along a highway with a velocity of 24 m/s, west. The car exits the highway; and 4.0 s later, its instantaneous vel
zloy xaker [14]

Answer:

4.25 m/s^{2}

Explanation:

Change in velocity considering the x component will be

Final velocity-Initial velocity

\triangle v_x= 16cos 45^{\circ}-24=-12.6862915 m/s

Change in velocity considering the y component will be

Final velocity-Initial velocity

\triangle v_y= 16sin 45^{\circ}-0=11.3137085 m/s

Resultant change in velocity=\sqrt {(-12.6862915 m/s)^{2}+(11.3137085 m/s)^{2}}=16.9982938 m/s

Acceleration= change in velocity per unit time hence

a= \frac {16.9982938}{4}=4.24957345\approx 4.25 m/s^{2}

5 0
3 years ago
an athlete whirls an 8.71 kg hammer tied to the end of a 1.5 m chain in a simple horizontal circle where you should ignore any v
-BARSIC- [3]

Answer:

T = 692.42 N

Explanation:

Given that,

Mass of hammer, m = 8.71 kg

Length of the chain to which an athlete whirls the hammer, r = 1.5 m

The angular sped of the hammer, \omega=1.16\ rev/s=7.28\ rad/s

We need to find the tension in the chain. The tension acting in the chain is balanced by the required centripetal force. It is given by the formula as follows :

F=m\omega^2r\\\\=8.71\times (7.28)^2\times 1.5\\\\=692.42\ N

So, the tension in the chain is 692.42 N.

5 0
3 years ago
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