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marusya05 [52]
3 years ago
8

Pirates drag a treasure chest to the left across a sandy beach. In which direction does the treasure chest experience a friction

al force, and why?
Physics
1 answer:
yawa3891 [41]3 years ago
5 0

Answer:

Chest experiences the friction forces towards right

Explanation:

Friction forces can be defined as a force a body experiences when it is made to slide against the surface. Friction force always acts to stop the the body which is moving. To stop the body, it always acts in the opposite directions to the motion of the body. Therefore, if the treasure chest is dragged across a sandy beach to the left, the frictional forces will act in the right direction.

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Which roots are better underwater , taprroots or fibrous?
4vir4ik [10]

Answer:

The absorption of water and minerals by taproot is more efficient with the taproot system. Fibrous root absorbs water more efficiently as it reaches deep into the soil. Taproots are capable of withstanding drought.

4 0
3 years ago
What is the magnitude of the electric field at the point if the electric potential in the region is given by V 2.00xyz2, where V
Hatshy [7]

Answer:

Electric field at a point ( x , y , z) is E=-2yz^2-2xz^2-4xyz .

Explanation:

Given :

Electric potential in the region is , V = 2xyz^2\ .

We need to find the electric field .

We know , electric field , E=-\dfrac{dV}{dr}  { Here r is distance }

In coordinate system ,

E=-\dfrac{dV}{\delta x }-\dfrac{dV}{\delta y }-\dfrac{dV}{\delta z }  { \delta is partial derivative }

Putting all values we get ,

E=-\dfrac{2xyz^2}{\delta x }-\dfrac{2xyz^2}{\delta y }-\dfrac{2xyz^2}{\delta z }\\\\E=-2yz^2-2xz^2-4xyz

Hence , this is the required solution.

3 0
3 years ago
A man in a gym is holding an 8.0-kg weight at arm's length, a distance of 0.55 m from his shoulder joint. What is the torque abo
kipiarov [429]

Answer:

  \tau =37.34\ N m

Explanation:

given,

mass of the weight = 8 Kg

distance = 0.55 m

angle below horizontal = 30°

torque about shoulder

  \tau = \vec{r} \times \vec{F}

  \tau = r \times F cos \theta

  \tau = 0.55 \times 8 \times 9.8 \times cos 30^0

  \tau =37.34\ N m

torque about his shoulder join is equal to   \tau =37.34\ N m

5 0
3 years ago
A person drives a car around a circular road with a constant speed of 20 m/s. The
ale4655 [162]

Answer:

16 m/s^2

Explanation:

acceleration tangential = (v^2)/r

a=400/25

a=16 m/s^2

Side note: next time, be more specific when asking about acceleration in circular motion. There's more than one type! Example:

angular acceleration=acceleration tangential/r

angular acc.=16/25

angular acc.=0.64 rad/s^2

5 0
3 years ago
Read 2 more answers
How much force is needed to accelerate a 68 kilogram skier at a rate of 1.2m/sec
xxMikexx [17]

Let's use Newton's 2nd law of motion:

                                     Force = (mass) x (acceleration)

                                     Force = (68 kg) x (1.2 m/s²) =  81.6 newtons .

8 0
3 years ago
Read 2 more answers
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