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STatiana [176]
2 years ago
7

When a 10-kg block is simultaneously pushed toward the east with 20 n and toward the west with 15 n, the combination of these fo

rces on the block is?
Physics
1 answer:
uysha [10]2 years ago
8 0

When a 10-kg block is simultaneously pushed toward the east with 20N and toward the west with 15N, the combination of these forces on the block is 5N toward the East

Let's understand this problem in detail

Let F_{1} = 20N which is in east direction

and F_{2} = 15N which is acting in the west direction

Now, F_{net} = F_{1} + F_{2}

But F_{1} and F_{2} are in opposite direction.

So let's rotate the force vector F_{2} by 180 degrees which becomes -F_{2}

Now both F_{1} and F_{2} are in same directionof the forces are in a single line

Adding both of them, we get

F_{net} = F_{1} + (-F_{2})

F_{net} = 20N - 15N = 5N, as both of the forces are in east direction hence, the net force is also acting in the east driection

Learn more about forces here

brainly.com/question/13191643

#SPJ4

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A rock is thrown off a cliff at an angle of 53° with respect to the horizontal. The cliff is 100 m high. The initial speed of th
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(a) 129.3 m

The motion of the rock is a projectile motion, consisting of two indipendent motions along the x- direction and the y-direction. In particular, the motion along the x- (horizontal) direction is a uniform motion with constant speed, while the motion along the y- (vertical) direction is an accelerated motion with constant acceleration g=-9.8 m/s^2 downward.

The maximum height of the rock is reached when the vertical component of the velocity becomes zero. The vertical velocity at time t is given by

v(t) = v_0 sin \theta +gt

where

v_0 = 30 m/s is the initial velocity of the rock

\theta=53^{\circ} is the angle

t is the time

Requiring v(t)=0, we find the time at which the heigth is maximum:

0=v_0 sin \theta + gt\\t=\frac{-v_0 sin \theta}{g}=-\frac{(30)(sin 53^{\circ})}{(-9.8)}=2.44 s

The heigth of the rock at time t is given by

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

Where h=100 m is the initial heigth. Substituting t = 2.44 s, we find the maximum height of the rock:

y=100+(30)(sin 53^{\circ})(2.44)+\frac{1}{2}(-9.8)(2.44)^2=129.3 m

(b) 44.1 m

For this part of the problem, we just need to consider the horizontal motion of the rock. The horizontal displacement of the rock at time t is given by

x(t) = (v_0 cos \theta) t

where

v_0 cos \theta is the horizontal component of the velocity, which remains constant during the entire motion

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If we substitute

t = 2.44 s

Which is the time at which the rock reaches the maximum height, we find how far the rock has moved at that time:

x=(30)(cos 53^{\circ})(2.44)=44.1 m

(c) 7.58 s

For this part, we need to consider the vertical motion again.

We said that the vertical position of the rock at time t is

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

By substituting

y(t)=0

We find the time t at which the rock reaches the heigth y=0, so the time at which the rock reaches the ground:

0=100+(30)(sin 53^{\circ})t+\frac{1}{2}(-9.8)t^2\\0=100+23.96t-4.9t^2

which gives two solutions:

t = -2.69 s (negative, we discard it)

t = 7.58 s --> this is our solution

(d) 136.8 m

The range of the rock can be simply calculated by calculating the horizontal distance travelled by the rock when it reaches the ground, so when

t = 7.58 s

Since the horizontal position of the rock is given by

x(t) = (v_0 cos \theta) t

Substituting

v_0 = 30 m/s\\\theta=53^{\circ}

and t = 7.58 s we find:

x=(30)(cos 53^{\circ})(7.58)=136.8 m

(e) (36.1 m, 128.3 m), (72.2 m, 117.4 m), (108.3 m, 67.4 m)

Using the equations of motions along the two directions:

x(t) = (v_0 cos \theta) t

y(t) = h+(v_0 sin\theta) t + \frac{1}{2}gt^2

And substituting the different times, we find:

x(2.0 s)=(30)(cos 53^{\circ})(2.0)=36.1 m

y(2.0 s)= 100+(23.96)(2.0)-4.9(2.0)^2=128.3 m

x(4.0 s)=(30)(cos 53^{\circ})(4.0)=72.2 m

y(4.0 s)= 100+(23.96)(4.0)-4.9(4.0)^2=117.4 m

x(6.0 s)=(30)(cos 53^{\circ})(6.0)=108.3 m

y(6.0 s)= 100+(23.96)(6.0)-4.9(6.0)^2=67.4 m

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