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Misha Larkins [42]
3 years ago
5

Two steamrollers begin 100 m apart and head toward each other, each at a constant speed of 1.00 m/s . At the same instant, a fly

that travels at a constant speed of 2.40 m/s starts from the front roller of the southbound steamroller and flies to the front roller of the northbound one, then turns around and flies to the front roller of the southbound once again, and continues in this way until it is crushed between the steamrollers in a collision.
What distance does the fly travel?
Physics
1 answer:
Debora [2.8K]3 years ago
4 0

Answer:

The fly travels 2.4 m

Explanation:

Since the Two steamrollers begin 100 m apart and head toward each other, each at a constant speed of 1.00 m/s, we can find the time until they crash by the formula:

Distance = Speed × Time

Time = Distance /Speed

Time = (100 m) / (1 m/s)

Time = 100 hours

Now, the fly will spend the same amount of time traveling as the steamrollers.

Since the fly moves at a speed of 2.4 m/s and we have a time of one hour the steamroller take to collide, then the fly will go a distance of;

Distance = speed x time = 2.4 × 1 = 2.4 m

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Colt1911 [192]

Answer:

Gravity is one major force that creates tides. In 1687, Sir Isaac Newton explained that ocean tides result from the gravitational attraction of the sun and moon on the oceans of the earth (Sumich, J.L., 1996).

Explanation:

I hope this helps.

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2 years ago
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The equations are not balanced. Which equation would have the same coefficients in the same order as 2CO2 + 3H20 → C2H6O + 3O2?
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<span>A: Al + FeO → Al2O3 + Fe

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5 0
3 years ago
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An automobile with a tangential speed of 54.1 km/h follows a circular road that has a radius of 41.6 m. The automobile has a mas
Viefleur [7K]

1) Available force of friction: 6174 N

2) No

Explanation:

1)

The magnitude of the frictional force between the car's tires and the pavement of the road is given by

F_f=\mu mg

where

\mu is the coefficient of friction

m is the mass of the car

g is the acceleration of gravity

For the car in this problem, we have:

\mu=0.500 (coefficient of friction)

m = 1260 kg (mass of the car)

g=9.8 m/s^2

Therefore, the force of friction is

F_f=(0.500)(1260)(9.8)=6174 N

2)

In order to mantain the car in circular motion, the force of friction must be at least equal to the centripetal force.

The centripetal force is given by

F=m\frac{v^2}{r}

where

m is the mass of the car

v is the tangential speed

r is the radius of the curve

In this problem, we have

m = 1260 kg

v=54.1 km/h =15.0 m/s is the tangential speed

r = 41.6 m is the radius of the curve

Therefore, the centripetal force is

F=(1260)\frac{15.0^2}{41.6}=6814 N

Therefore, the force of friction is not enough to keep the car in the curve, since F_f

4 0
3 years ago
A 4kg ball moving at 8m\sec
Llana [10]
Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32

Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32

Therefore β = 12.8 m/s
8 0
3 years ago
A 398-kg boat is sailing 14.0° north of east at a speed of 1.50 m/s. 23.0 s later, it is sailing 33.0° north of east at a speed
kicyunya [14]

Answer:

Explanation:

We shall represent speed in vector form

First speed

v₁ = 1.5 cos 14 i + 1.5 sin 14 j

= 1.455 i + 0.363 j

v₂ = 4.4 cos 33 i + 4.4 sin 33 j

= 3.69 i + 2.39 j

v₂ - v₁

3.69 i + 2.39 j - 1.455 i - 0.363 j

= 2.235 i + 2.027 j

acceleration

=  v₂ - v₁ / time

= ( 2.235 i + 2.027 j  ) / 23

= .097 i + .088 j

force = mass x acceleration

= 398 x ( .097 i + .088 j )

= 38.6 i + 35.02 j

Magnitude of force F

F² = 38.6² + 35.02²

F = 52.11 N

Tan θ = 35.02 / 38.6

θ = 42° north of east.

4 0
2 years ago
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