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10 months ago

8

Calculate the magnitude of the acceleration of the box if you push on the box with a constant force 170.0 N that is parallel to

the ramp surface and directed up the ramp, moving the box up the ramp.
Express your answer with the appropriate units.
a₁ = 6.5m²

1 answer:

Degger [83]10 months ago

8 0

The acceleration of the box up the ramp is 9.65 m/s².

<h3>What is the magnitude of acceleration of the box?</h3>

The magnitude of the acceleration of the box is calculated by applying Newton's second law of motion as shown below;

F(net) = ma

where;

m is the mass of the box
a is the acceleration of the box

The net force on the box is calculated as follows;

F(net) = F - Ff

F(net) = F - μmgcosθ

where;

θ is the inclination of the plane
μ is coefficient of friction

F(net) = 170 - (0.3 x 15 x 9.8 x cos55)

F(net) = 144.7

The acceleration of the box is calculated as;

a = F(net) / m

a = (144.7) / (15)

a = 9.65 m/s²

Thus, the acceleration of the box up the ramp is 9.65 m/s².

Learn more about acceleration here: brainly.com/question/14344386

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Answer:

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Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

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$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{v^2-0^2}{2\times h}\\\Rightarrow v^2=2ah\ m/s$

Now H-h = 42.3 - 0.2 = 42.1 cm = 0.421 m

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$v^2-u^2=2as\\\Rightarrow 0^2-u^2=2gs\\\Rightarrow -2ah=2\times g(H-h)\\\Rightarrow -2a0.002=2\times g0.421\\\Rightarrow a=-\frac{0.421\times -9.81}{0.002}\\\Rightarrow a=2065.005\ m/s^2$

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2 years ago

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$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

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Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

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$X=v_{ox}*t$

$v_{0x} =v_0cos(\alpha)$

Y axis:

$Y= Y_0 +v_{y0} t - \frac{g}{2} t^2$

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

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But, in physics, time it could not be negative, so we take $t_1= 10.18$

Caculate now:

$X=79m/s*cos(\39)*10.18s= 624.996 m$

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$v_0= 79m/s+13m/s= 92m/s$

Using the same procedure that item A, caculate X

First, we need to know the new time

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And we obtain:

$t_1=11.845s\\t_2=-0.041s$

One more time, we take the positive time: $t_1=11.845s$

Finally:

$X=92m/s *cos(39)*11.845s=846.887 m$

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Explanation:

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Answer:

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