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2 years ago

14

A weightlifter walks through 2 m, on his head with an object of 800 N that he has lifted . What is the total work done in walkin

g 2 m distance?

1 answer:

jeyben [28]2 years ago

4 0

Work = force x displacement
work = 800 N x 2 m
work = 1600 Joules

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A fox is 7 meters from the base of a tree that is 15 m tall. A nest of owls is at the top of the tree. One of the owlets tosses

mamaluj [8]

Refer to the diagram shown below.

In this analysis, wind resistance is ignored, and g = 9.8 m/s².

The meat falls with zero vertical velocity, therefore the time, t, before the meat hits the ground is
$\frac{1}{2}*(9.8 \, \frac{m}{s^{2}})*(t \, s)^{2} = (15 \, m) \\ t= \sqrt{ \frac{15}{4.9} }= 1.75 \, s$

If the fox catches the meat before it hits the ground, then the fox should travel a horizontal distance d in the same time that the meat travels a horizontal distance (7 -d).
The meat travels a distance of
7 - d = (1.2 m/s)*(1.75 s) = 2.1 m
or
d = 4.9 m

Let v = velocity of the fox when it catches the meat.
If the acceleration of the fox is a m/s², then
v = 1.75a
Also,
$d= \frac{1}{2} *(a \, \frac{m}{s^{2}} )*(1.75 \, s)^{2} = \frac{1}{2}( \frac{v}{1.75})^{2}*(1.75^{2}) \\ 4.9 = 0.875v^{2} \\ v^{2} = 5.6 \\ v = 2.366 \, m/s$

Answer: 2.37 m/s (nearest hundredth)

5 0

4 years ago

Three resistors, 21 Ω, 58 Ω, and 64 Ω, are connected in series, and a 0.50-A current passes through them. What are (a) the equiv

adoni [48]

<h2>a) Equivalent resistance is 143 Ω</h2><h2>b) Potential difference is 71.5 V</h2>

Explanation:

When resistors are connected in series, effective resistance is given by

$R_{eff}=R_1+R_2+R_3+......$

Here

R₁ = 21Ω

R₂ = 58Ω

R₃ = 64Ω

a) $R_{eff}=R_1+R_2+R_3\\\\R_{eff}=21+58+64\\\\R_{eff}=143\Omega$

Equivalent resistance is 143 Ω

b) We know

Potential difference = Current x Resistance

V = IR

I = 0.5 A

R = 143Ω

Substituting

V = 0.5 x 143 = 71.5 V

Potential difference is 71.5 V

6 0

4 years ago

(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from

victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c) $a_{satellite}= 3.13 m/sec^2$

d) $a_{earth}=2.32\times10^{-22} m/s^2$

Explanation:

The force experienced by the satelite is giveb by

$F= \frac{Gm_{satellite}m_{earth}}{r^2} \\$

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

$F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}$

⇒F= 1396.52 N

now,

$a_{satellite}= \frac{F}{m_{satellite}}$

$a_{satellite}= \frac{1396.52}{445}$

$a_{satellite}= 3.13 m/sec^2$

also,

$a_{earth}= \frac{F}{m_{earth}}$

$a_{earth}= \frac{1396.52}{(6\times10^24)}$

$a_{earth}=2.32\times10^{-22} m/s^2$

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4 years ago

A velocity vs. time graph starts at 0 and ends at 10 m/s, stretching over a time- span of 15 s with constant acceleration. What

Alex777 [14]

Answer:

75m

Explanation:

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4 years ago

What is the momentum of a 3 kg bowling ball moving at 3 m/s?

Nataly [62]

Explanation:

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<em>m</em> denotes mass
<em>v</em> denotes velocity

→ p = 3 kg × 3 m/s

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3 years ago