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2 years ago

Please help me, this is a physics test.

Physics

1 answer:

sweet-ann [11.9K]2 years ago

4 0

Answer:

a = 2 [m/s²]

Explanation:

To be able to solve this problem we must make it clear that the starting point when the time is equal to zero, the velocity is 5 [m/s] and when three seconds have passed the velocity is 11 [m/s], this point is the final point or the final velocity.

We can use the following equation.

$v_{f}=v_{o}+a*t\\$

where:

Vf = final velocity = 11 [m/s]

Vo = initial velocity = 5 [m/s]

a = acceleration [m/s²]

t = time = 3 [s]

$11 = 5 + a*3\\6=3*a\\a= 2[m/s^{2} ]$

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3 years ago

Read 2 more answers

An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce

ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

$v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}$

From them we can get $v^2=v_0^2+2ad$ .

We have:

$v-v_0=at\\t=\frac{v-v_0}{a}$

And substitute:

$x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}$

We multiply both sides by 2a, and continue:

$2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2$

Being d the displacement $x-x_0$ , we have $v^2=v_0^2+2ad$

For our exercise, we will write this as:

$d=\frac{v^2-v_0^2}{2a}$

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

$d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m$

For the time we use:

$t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s$

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Answer:

$1m/s^2$

Explanation:

Mass of block=10 kg

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