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3 years ago

What equation gives the position at a specific time for an object with constant acceleration?

1 answer:

4 0

Given constant acceleration, we can get the final position of an object in terms of both its initial velocity and its acceleration using one of the equations of motion.

The equation that we will use is:
Xf = Xi + Vi*t + (1/2)*a*t^2
where:
Xf is the final position of the object
Xi is the initial position of the object
Vi is the initial velocity of the object
t is the time
a is the constant given acceleration

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Help me with this PLEASE, I'M being timed

juin [17]

Answer:

Explanation:

6 0

3 years ago

: A 70 kg man and a 12 kg sled are on the frictionless ice of a frozen lake, 25 m apart but connected by a rope of negligible ma

e-lub [12.9K]

Answer:

$x_1 = 3.74m$

Explanation:

given,

mass of man = 70 kg

mass of sled = 12 kg

F = m a_s

$a_s = \dfrac{F}{m}$

$a_s = \dfrac{8.2}{12}$

$a_s = 0.68\ m/s^2$

$F = m a_m$

$a_m = \dfrac{F}{m}$

$a_m = \dfrac{8.2}{70}$

$a_m = 0.12\ m/s^2$

$x_1+x_2 = 25$

$\dfrac{1}{2}a_ct^2+ \dfrac{1}{2}a_mt^2 = 25$

$(a_c+a_m)t^2=50$

$(0.12+0.68)t^2=50$

$t = \sqrt{\dfrac{50}{0.8}}$

t = 7.90 s

$x_1 = \dfrac{1}{2}a_ct^2$

$x_1 = 0.5\times 0.12 \times 7.90^2$

$x_1 = 3.74m$

5 0

3 years ago

What do all wetlands have in common?

Ksivusya [100]

They are all coverd in water 24/7 they never clear up

4 0

3 years ago

Read 2 more answers

A 24.7-g bullet is fired from a rifle. It takes 2.73 × 10-3 s for the bullet to travel the length of the barrel, and it exits th

Gala2k [10]

Answer:

$F_a_v_g=7093333.33N*s$

Explanation:

The impulse or average force in classical mechanics is the variation in the linear momentum that a physical object experiences in a closed system. It is defined by the following equation:

$F_a_v_g=m*\frac{\Delta v}{\Delta t}=m*\frac{v_2-v_1}{t_2-t_1}$

Where:

$m=mass\hspace{3}of\hspace{3}the\hspace{3}object$

$v_2=final\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}at\hspace{3}the\hspace{3}end\hspace{3}of\hspace{3}the\hspace{3}time\hspace{3}interval$

$v_1=initial\hspace{3}velocity\hspace{3}of\hspace{3}the\hspace{3}object\hspace{3}when\hspace{3}the\hspace{3}time\hspace{3}interval\hspace{3}begins.$

$t_2=final\hspace{3}time$

$t_1=initial\hspace{3}time$

Asumming v1=0 and t1=0:

$F_a_v_g=m* \frac{v_2}{t_2} =(24.7)*\frac{784}{2.73*10^{*3} } =7093333.333N*s$

8 0

3 years ago

Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105

Dovator [93]

It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
All you need is F=(K*Q1*Q2)/r^2 
Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you 
(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C

3 0

3 years ago

Read 2 more answers