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Rom4ik [11]
3 years ago
9

1 point

Physics
1 answer:
Hitman42 [59]3 years ago
5 0
I believe it is Gravitational force, hope this helps
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A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in
katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= 10+1.50

= 11.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v = \frac{d}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{11.5}

      =26.08 \ s

Throughout the opposite direction, when the boat seems to be travelling then,

= 10-1.50

= 8.5 \ m/s

Travelling such distance for 300 m will be:

⇒ v=\frac{v}{t} \ sot \ t

      =\frac{d}{v}

On putting the values, we get

      =\frac{300}{8.5}

      =35.29 \ s

hence,

The time taken by the boat will be:

= 26.08+35.29

= 61.37 \ s

8 0
3 years ago
How would the plane strain fracture toughness of a metal be expected to change with rising temperature?
aksik [14]

Answer:

Increase

Explanation:

The plane strain fracture toughness of a metal is expected to increase with rising temperature.

4 0
3 years ago
A ball is projected upward at time t = 0.0 s, from a point on a roof 60 m above the ground. The ball rises, then falls until it
musickatia [10]
<span> y=y0 + vt +1/2gt^2
(solve for t here) cause you know y,y0,v,g
you will do quad formula here

then:
v=v0 +at solve for v
(remember the direction of the ball too (signs))

The main thing to remember here is that when the ball passes exactly (height) where it was launched it will travel the speed at which it was launched. *its almost like the ball was thrown in the downward direction. </span>
7 0
3 years ago
A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel
trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

1\ mile=1609.34\ m

1\ h=3600\ seconds

18.03\times \frac{3600}{1609.34}=40.33\ mph

Speed of the boat by when it will hit the dock is 40.33 mph

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m

The distance at which the boat will have to start decelerating is 312.5 m

5 0
3 years ago
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
3 years ago
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