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3 years ago

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A boxed 12.0 kg computer monitor is dragged by friction 5.50 m up along the moving surface of a conveyor belt inclined at an ang

le of 36.4 ∘ above the horizontal. If the monitor's speed is a constant 3.00 cm/s , how much work is done on the monitor by friction, gravity, and the normal force of the conveyor belt

1 answer:

satela [25.4K]3 years ago

5 0

Nah a a yuloure

Explanation:

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A cheetah can run at a maximum speed 97.8 km/h and a gazelle can run at a maximum speed of 78.2 km/h. If both animals are runnin

emmainna [20.7K]

(1)

Cheetah speed: $v_c = 97.8 km/h=27.2 m/s$

Its position at time t is given by

$S_c (t)= v_c t$ (1)

Gazelle speed: $v_g = 78.2 km/h=21.7 m/s$

the gazelle starts S0=96.8 m ahead, therefore its position at time t is given by

$S_g(t)=S_0 +v_g t$ (2)

The cheetah reaches the gazelle when $S_c=S_g$ . Therefore, equalizing (1) and (2) and solving for t, we find the time the cheetah needs to catch the gazelle:

$v_c t=S_0 + v_g t$

$(v_c -v_g t)=S_0$

$t=\frac{S_0}{v_c-v_t}=\frac{96.8 m}{27.2 m/s-21.7 m/s}=17.6 s$

(2) To solve the problem, we have to calculate the distance that the two animals can cover in t=7.5 s.

Cheetah: $S_c = v_c t =(27.2 m/s)(7.5 s)=204 m$

Gazelle: $S_g = v_g t =(21.7 m/s)(7.5 s)=162.8 m$

So, the gazelle should be ahead of the cheetah of at least

$d=S_c -S_g =204 m-162.8 m=41.2 m$

4 0

3 years ago

Read 2 more answers

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

Ilya [14]

Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

$t = 0.192 \mu m$

3 0

3 years ago

I'm not an idiot so I'm pretty sure c and d are not the answers. I'm confused because don't they both do this. As it picks up sp

krok68 [10]

The answer is A because it's MOVING

ithink

8 0

3 years ago

Read 2 more answers

What 2 environments once existed where the Grand Canyon is located today?

patriot [66]

The entire park area is considered to be a semi-arid desert, but distinct habitats are located at different elevations along the 8,000-foot elevation gradient. Near the Colorado River, riparian vegetation and sandy beaches prevail.

5 0

3 years ago

3.00 m^3 of water is at 20.0°C.

krok68 [10]

Answer:

9m^3

Explanation:

Given data

volume v1= 3m^3

volume v2= ???

Temperature T1= 20.0°C.

Temperature T2= 60.0°C.

Applying the relation for temperature and volume

V1/T1= V2/T2

substitute

3/20= V2/60

3*60= V2*20

180= 20*V2

180/20= V2

V2= 9m^3

Hence the final volume is 9m^3

6 0

3 years ago