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Tom [10]
2 years ago
7

2 someone was described in a novel as riding off into the sunset," then they'd be travelling west. o True o False ​

Physics
1 answer:
Vanyuwa [196]2 years ago
5 0

Answer: west

Explanation: the sunset always sets in the west.

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Explanation for x=73km/h•0.5s= 10.14 pls help how did they get 10.14? What was the work around it
Sergio [31]
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4 0
3 years ago
A 60.0 kg person walks from the ground to the roof of a 74.8 m tall building.how much potential energy doed the person have at t
faust18 [17]
So first, we multiply mass by gravity by height, which is 60 * 9.8 * 74.8 = <span>43747.2

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5 0
3 years ago
A block of wood measuring 2.75 cm x 4.80 cm x 7.50 cm has a mass of 84.0 g. will the block of wood sink or float in water?
KIM [24]
The volume of the block of wood is given by length × width ×height
 = 2.75 × 4.80 × 7.5
 = 99 cm³
Density is given by mass/volume
Thus = 84.0 g/ 99 cm³
         = 0.848 g/cm³
Hence;  since the block is less dense than water (1 g/cm³) it will float
6 0
3 years ago
A wheel rotates about a fixed axis with an initial angular velocity of 24 rad/s. During a 4 s interval the angular velocity decr
netineya [11]

Answer:

\theta=76\ rad

Explanation:

Hoven that,

Initial angular velocity of the wheel = 24 rad/s

Final angular velocity = 14 m/s

Time, t = 4 s

We need to find how many radians does the wheel turn through during the 4 s interval. Let the displacement is \theta. Using second equation of rotational kinematics to find it such that,

\theta=\omega_i t+\dfrac{1}{2}\alpha t^2

Where

\alpha is angular acceleration

\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{14-24}{4}\\\\\alpha =-2.5\ rad/s^2

So,

\theta=24\times 4+\dfrac{1}{2}\times (-2.5)\times 4^2\\\\\theta=76\ rad

So, it will turn 76 radian during the 4 s interval.

7 0
3 years ago
You place a box weighing 289.9 N on an inclined plane that makes a 35.9 degree angle with the horizontal. Compute the component
Ksivusya [100]

Answer:

Fgparallel =  170N

Explanation:

Acting down the incline would be the paralell force. The Force of gravity on an incline for the parallel portion is mgsinθ.

Fg parallel = mgsinθ

mg is 289.9, as that is the weight. θ is 35.9

Fgpar = 289.9sin35.9

Fgparallel =  170N

5 0
3 years ago
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