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masha68 [24]
2 years ago
7

What is the answer to this question?

Physics
1 answer:
icang [17]2 years ago
8 0

Answer:

Explanation:

In a velocity/time (aka acceleration) graph, the slope of a line indicates the value of the acceleration in m/s/s. Acceleration is the change in velocity over the change in time. From 0 - 2 seconds, there is no change in velocity, so the acceleration during this interval is 0 (which is the same as the slope of the line). From 2 - 4 seconds, the slope of the line is -2, so the acceleration during the time interval from 2 to 4 seconds is -2 (negative because David is slowing down but is still going the same direction: to the right).

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2. Consider a conical pendulum with a bob of mass m = 77.0 kg on a string of length L = 10.0 m that makes an angle of ???? 3.00°
bekas [8.4K]

Explanation:

It is given that,

The mass of bob, m = 77 kg

Length of the string, L = 10 m

Angle made by the string with the vertical, \theta=3^{\circ}

(a) Let T is the force exerted by the string on the pendulum. At equilibrium,

T\ cos\theta=mg

T=\dfrac{mg}{cos\theta}

T=\dfrac{77\times 9.8}{cos(3)}

T = 755.63 N

The horizontal component of the force is given by,

T_H=T\ sin\theta

T_H=755.63\ sin(3)=39.54\ N

The vertical component of the force is given by,

T_V=T\ sin\theta=mg

T_V=754.59\ N

(b) Let a is the radial acceleration of the bob. It can be calculated as :

a=\dfrac{T_H}{m}

a=\dfrac{39.54\ N}{77\ kg}

a=0.51\ m/s^2

Hence, this is the required solution.

6 0
2 years ago
Finally, you are ready to answer the main question. Cheetahs, the fastest of the great cats, can reach 50.0 miles/hour miles/hou
dolphi86 [110]

Answer:

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

Explanation:

Acceleration is the change in velocity per unit time

a = ∆v/t

Given;

∆v = 50.0miles/hour - 0

∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour

∆v = 22.352m/s

t = 2.22 s

So,

Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s

a = 10.07m/s^2

Their acceleration in meters per second squared is 10.07m/s^2

7 0
2 years ago
Read 2 more answers
A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin
ollegr [7]
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
8 0
3 years ago
When doing scientific research ,the sources used should be ?
quester [9]

Answer: Reliable and trusted

6 0
3 years ago
Read 2 more answers
Two circular coils are concentric and lie in the same plane. The inner coil contains 170 turns of wire, has a radius of 0.0095 m
PIT_PIT [208]

Answer:

Current in outer circle will be 15.826 A

Explanation:

We have given number of turns in inner coil N_I=170

Radius of inner circle r_i=0.0095m

Current in the inner circle I_i=8.9A

Number of turns in outer circle N_o=150

Radius of outer circle r_o=0.015m

We have to find the current in outer circle so that net magnetic field will zero

For net magnetic field current must be in opposite direction as in inner circle

We know that magnetic field is given due to circular coil is given  by

B=\frac{N\mu _0I}{2r}

For net magnetic field zero

\frac{N_I\mu _0I_I}{2r_I}=\frac{N_O\mu _0I_0}{2r_O}

So \frac{170\times \mu _0\times 8.9}{2\times 0.0095}=\frac{150\times \mu _0I_O}{2\times 0.015}

I_O=15.92A

4 0
2 years ago
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