What is the speed of a bobsled whose distance-time graph indicates that it traveled 123m in 28s?

If the population runs out of food after 6 hours, what would most likely happen to the graph line between 8 and 10 hours?

padilas [110]

It would decrease because there is no more food

3 0

2 years ago

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A 30kg uniform solid cylinder has a radius of 0.18m. if the cylinder accelerates at 0.023 rad/s^2 as it rotates about an axis th

Nuetrik [128]

Answer:

0.011 N-m

Explanation:

Given that

The mass of a solid cylinder, m = 30 kg

The radius of the cylinder, r = 0.18 m

The acceleration of the cylinder, $\alpha =0.023\ rad/s^2$

It rotates about an axis through its center. We need to find the torque acting on the cylinder. The formula for the torque is given by :

$\tau=I\alpha$

Where

I is the moment of inertia of the cylinder,

For cylinder,

$I=\dfrac{mr^2}{2}$

So,

$\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{30\times (0.18)^2\times 0.023 }{2}\\\\\tau=0.011\ N-m$

So, the required torque on the cylinder is 0.011 N-m.

7 0

2 years ago

What is the potential difference across lamp 1

Alex73 [517]

Answer:

1+1=2

Explanation:

You just need to add and voila the answer

7 0

2 years ago

Which electromagnetic wave is used in dental scans?

julsineya [31]

Answer:

X-rays are most often used to examine bones and teeth.

Hope this helps!

8 0

3 years ago

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A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1

Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = $\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]$

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = $\frac{\rho dx}{A}$

$dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}$

Taking the integral of both sides;we have:

$\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx$

$R = 3.185 [x + \frac {x^3}{3}}]^2__0$

$R = 3.185 [2 + \frac {2^3}{3}}]$

R = 14.863 Ω

Since V = IR

$I = \frac{V}{R}$

$I = \frac{17}{14.863}$

I = 1.144 A

∴ the current if this wire if it is connected to the terminals of a 17.0V battery = 1.144 A

8 0

3 years ago