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3 years ago

7

A man is chased by a bull in a field at a speed of 10m/s for 60 seconds.How far does he get before he is gored by the annoyed bo

vine?

1 answer:

Vsevolod [243]3 years ago

5 0

60 times 10 =600 well if u multiply 60 times 10 it will=600

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Where do you think the government should put the greatest support: solar energy, wind energy, clean coal, oil exploration, gas e

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The government should put it support in a combination of sources, as no source in the present can fully provide all energy requirements.

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2 years ago

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Two mass m1 and m2 lie on a frictionless surface. Between the two masses is a compressed spring, with spring constant k. The sys

max2010maxim [7]

Answer:

The spring was compressed the following amount:

$\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

Explanation:

Use conservation of energy between initial and final state, considering that the surface id frictionless, and there is no loss in thermal energy due to friction. the total initial energy is the potential energy of the compressed spring (by an amount $\Delta x$ ), and the total final energy is the addition of the kinetic energies of both masses:

$E_i=\frac{1}{2} k\,(\Delta x)^2\\\\E_f=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2$

$E_i=E_f\\$

$\frac{1}{2} k\,(\Delta x)^2=\frac{1}{2} m_1\,v_1^2+\frac{1}{2} m_2\,v_2^2\\k\,(\Delta x)^2=m_1\,v_1^2+ m_2\,v_2^2\\(\Delta x)^2=\frac{m_1\,v_1^2+ m_2\,v_2^2}{k} \\\Delta x=\sqrt{ \frac{m_1\,v_1^2+ m_2\,v_2^2}{k} }$

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2 years ago

A 7300 N elevator is to be given an acceleration of 0.150g by connecting it to a cable of negligible weight wrapped around a tur

Firlakuza [10]

Answer:

α =18.75 rad/s²

Explanation:

Given that

Acceleration a = 0.15 g

We know that g =10 m/s²

a= 0.15 x 10 = 1.5 m/s²

d= 16 cm

Radius r= 8 cm

Lets take angular acceleration =α rad/s²

As we know that

a= α r

Now by putting the values

1.5 = α x 0.08

α =18.75 rad/s²

6 0

3 years ago

Pls help answer embed <br>

Savatey [412]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, $v_t=60\ m/s$

Area of cross section, $A=0.33\ m^2$

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

Weight of the object = drag force

R = W

or

$\dfrac{1}{2}\rho CAv_t^2=mg$

Where

$\rho$ is the density of air = 1.225 kg/m³

C is drag coefficient

So,

$C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01$

So, the drag coefficient is 1.01.

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2 years ago

Polar molecules have _____. A) an uneven distribution of charges around the molecule B) atoms that are the same element C) an ev

Keith_Richards [23]

Answer: ionic bonds

Explanation:

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2 years ago

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