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2 years ago

Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

Physics

1 answer:

konstantin123 [22]2 years ago

5 0

Answer: His average speed in mph over the last 400 m is 7.7 m/s.

Explanation:

Given: Hicham El Guerrouj of Morocco holds the world record in the 1500 m running race. He ran the final 400 m in a time of 51.9 s.

We know that , speed = $\dfrac{distance}{time}$

Here , distance = 400m and time = 51.9 s

Then, speed = $\dfrac{400}{51.9}\approx7.7\ m/s$

Hence, his average speed in mph over the last 400 m is 7.7 m/s.

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A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

The magnetic field at the earth's surface can vary in response to solar activity. During one intense solar storm, the vertical c

Flauer [41]

Answer:

$EMF = 33880 Volts$

Explanation:

As per Faraday's law of Electromagnetic induction we know that

Rate of change in magnetic flux will induce EMF in the closed conducting loop

so we have

$\phi = B.A$

now we have

$A = (110 \times 10^3)(110 \times 10^3)$

$A = 1.21 \times 10^{10}$

now we have

$\phi = B(1.21 \times 10^{10})$

now the induced EMF through this loop is given as

$EMF = (\frac{dB}{dt})(1.21 \times 10^{10})$

$EMF = (2.8 \times 10^{-6})(1.21 \times 10^{10})$

$EMF = 33880 Volts$

5 0

2 years ago

A stone is thrown vertically upward with a speed of 15.0 m/s from the edge of a cliff 75.0 m high.How much later does it reach t

Katarina [22]

Answer:

5.72 seconds

848.27 m/s

97.94 m

Explanation:

t = Time taken

u = Initial velocity = 15 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=15-9.81\times t\\\Rightarrow \frac{-15}{-9.81}=t\\\Rightarrow t=1.52 \s$

Time taken to reach maximum height is 0.97 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=15\times 1.52+\frac{1}{2}\times -9.81\times 1.52^2\\\Rightarrow s=11.47\ m$

So, the stone would travel 11.47 m up

So, total height stone would fall is 75+11.47 = 86.47 m

Total distance travelled by the stone would be 75+11.47+11.47 = 97.94 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 86.47=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{86.47\times 2}{9.81}}\\\Rightarrow t=4.2\ s$

Time taken by the stone to travel 86.47 m to the water is is 4.2 seconds

The stone reaches the water after 4.2+1.52 = 5.72 seconds after throwing the stone

$v=u+at\\\Rightarrow v=0+9.81\times 86.47 = 848.27\ m/s$

Speed just before hitting the water is 848.27 m/s

3 0

2 years ago

A 45-mH inductor is connected to an ac source of emf with a frequency of 250 Hz and a maximum emf of 20 V. If the voltage across

ruslelena [56]

Answer:

-20v

Explanation:

Data provided in the question as per the given situation

L = 45 mH

f = 250 Hz

V_o = 20V

Based on the above information, let us assume voltage across inductor be V at t

So,

$V = 20 sin (2 \pi \times 250\times t + \frac{\pi}{2})$

t = 2 ms

Now

The voltage at time t = 2.0 ms is

$V_1 = 20 sin (500\times 2\times 10^{-1} \times \pi + \frac{\pi}{2})$

$= 20 sin(\pi + \frac{\pi}{2} )$

= -20 volt

= -20v

We simply solved the above equation so that the correct voltage could come

5 0

2 years ago

) a charge of 6.15 mc is placed at each corner of a square 0.100 m on a side. determine the magnitude and direction of the force

Nana76 [90]

Because charges are positioned on a square the force acting on one charge is the same as the force acting on all others.
We will use superposition principle. This means that force acting on the charge is the sum of individual forces. I have attached the sketch that you should take a look at.
We will break down forces on their x and y components:
$F_x=F_3+F_2cos(45^{\circ})$
$F_y=F_1+F_2sin(45^{\circ})$
Let's figure out each component:
$F_1=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\
F_3=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}\\$
$F_2=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}$
Total force acting on the charge would be:
$F=\sqrt{F_x^2+F_y^2}$
We need to calculate forces along x and y axis first( I will assume you meant micro coulombs, because otherwise we get forces that are huge).
$F_x=F_3+F_2cos(45^{\circ})=\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}+\frac{1} {4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot\cos(45)=46N$
$F_y=\frac{1}{4\pi \epsilon}\frac{q^2}{(\sqrt{2}a)^2}\cdot sin(45)+\frac{1}{4\pi \epsilon}\frac{q^2}{a^2}=46N$
Now we can find the total force acting on a single charge:
$F=\sqrt{F_x^2+F_y^2}=\sqrt{46^2+46^2}=65N$
As said before, intensity of the force acting on charges is the same for all of them.

5 0

2 years ago