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Svetach [21]
3 years ago
5

A horse starts from rest and speeds uo with constant acceleration. If it has gone a distance of 100 m at the point when it reach

es a speed of 12 m/s, what is the acceleration?
Physics
1 answer:
makkiz [27]3 years ago
3 0

Answer:

yee yee

Explanation:

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If Vx = 7.00 units and Vy = -7.60 units, determine the magnitude of V⃗ .
Juliette [100K]

|V| = 10.33 units and the direction θ = -47.35° or 312.65°.

Given the x and y components of a vector, we can calculate the magnitude and direction from these components.

Applying the Pythagorean theorem we have that the magnitude of the vector is:

|V| = \sqrt{Vx^{2}+Vy^{2}  }

|V| = \sqrt{(7.00units)^{2}+(-7.60units)^{2}} = \sqrt{106units^{2}} = 10.33units

The expression for the direction of a vector comes from the definition of the tangent of an angle:

tan θ = \frac{Vy}{Vx} ------>  θ = arc tan \frac{Vy}{Vx}

θ = arc tan \frac{-7.60units}{7.00units}

θ = -47.35° or 312.65°

6 0
3 years ago
A uniform metal tube of length 5m and mass 9kg is suspended by two vertical wires attached at 50cm and 150cm respectively from t
Elena-2011 [213]

Answer:

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

Explanation:

Given:

Length of tube = 5 m (500 cm)

Mass of tube = 9

Suspended vertically from 150 cm and 50 cm.

Computation:

Force = Mass × gravity acceleration.

Force = 9.8 x 9

Force = 88.2 N

So,

Upward forces = Downward forces

D1 = 150 - 50 = 100 cm

D2 = 150 + 50 = 200 cm

And F1 = F2

F1 x D1 = F2 x D2

F1 x 100 = F2 x 200

F = 2F

Total force = Upward forces + Downward forces

3F = 88.2

F = 29.4 and 2F = 58.8 N

force (tension) of 29.4 N (upward)  in 100 cm

force (tension) of 58.4 N (upward)  in 200 cm

4 0
4 years ago
explain why the EMF of a dry cell drops if a large current is drawn for a short time and then recovers if allowed to rest​
Delvig [45]

Answer:

Manganese (iv )oxide is a slow depolarizer and polarization occurs with a large current, on resting, the depolarization returns the p.d of the dry cell.

8 0
3 years ago
Calculate the acceleration of a skier heading down a 10.0º slope, assuming the coefficient of friction for waxed wood on wet sno
lutik1710 [3]

(a) 0.74 m/s^2

Explanation:

There are two forces acting on the skier: the component of the weight parallel to the slope, which acts downward, and the frictional force, which acts upward along the incline.

The component of the weight parallel to the inclined plane is:

W= m g sin \theta

where m is the mass of the skier, g=9.81 m/s^2 and \theta=10^{\circ}.

The frictional force is instead

F_f = -\mu m g cos \theta

\mu=0.1 is the coefficient of friction for waxed wood on wet snow.

If we apply Newton's second law, we can write that the net force must be equal to the product of mass per acceleration:

mgsin \theta -\mu mg cos \theta =ma

And symplifying m, we can find the acceleration:

a=g sin \theta-\mu g cos \theta=

=(9.81 m/s^2)(sin 10^{\circ})-(0.1)(9.81 m/s^2)(cos 10^{\circ})=0.74 m/s^2


(b) 5.7^{\circ}

Explanation:

This time, the skier is moving at constant velocity. Therefore, the acceleration is zero (a=0) and Newton's second law becomes:

mg sin \theta - \mu m g cos \theta=0

By simplifying, we get

tan \theta = \mu

From which we can find the angle at which the skier could coast at a constant velocity:

\theta= tan^{-1} (\mu) = tan^{-1} (0.1)=5.7^{\circ}



6 0
3 years ago
Two tougboats are toeing a ship each exerts a force of 6000N and the angle between the two ropes is 60 calculate the resultant f
Arlecino [84]

Answer:

10392.30N

Explanation:

We proceed by computing the individual force exerted by the boats

For the first boat

The angle is 30 degree to the vertical

Hence

Force = F cos θ

F=6000 cos 30

F=6000*0.866

F=5196.15 N

Since the boats are two and also at the same angle and also exerting the same force

The Net force = 2*5196.15

Net force=10392.30N

7 0
3 years ago
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