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1 year ago

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A constant braking force of 10 newtons applied for 5 seconds is used to stop a 2.5-kilogram cart traveling at 20 meters per seco

nd. the magnitude of the impulse applied to stop the cart is

2 answers:

ss7ja [257]1 year ago

6 0

Mathematically it can be expressed as: I = FpΔt. Where F p is the average magnitude of the acting force and Δt = t 2 - t 1, the time lapse in the force acts.
The magnitude of the impulse applied to stop the cart is
I = FpΔt = (10N) * (5s) = 50 N.s

inn [45]1 year ago

4 0

Answer:The magnitude of the impulse applied to stop the cart is 50 N sec.

Explanation:

Force applied on the cart = 10 N

Interval of time till force applied ,Time= 5 second

$Impulse=Force\times Time=10 N\times 5 seconds= 50$ N seconds

The magnitude of the impulse applied to stop the cart is 50 N sec.

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An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i

Butoxors [25]

Answer: $5 m/s^{2}$

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

$y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2}$ (1)

Where:

$y=0m$ is the rock's final height

$y_{o}=100 m$ is the rock's initial height

$V_{o}=15 m/s$ is the rock's initial velocity

$\theta=90\°$ is the angle at which the rock was thrown (directly upwards)

$t=10 s$ is the time

$g$ is the acceleration due gravity in Planet X

Then, isolating $g$ and taking into account $sin(90\°)=1$ :

$g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t)$ (2)

$g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s))$ (3)

Finally:

$g=5 m/s^{2}$ (4) This is the acceleration due gravity in Planet X

7 0

1 year ago

Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache

Nina [5.8K]

Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

R = v₀² sin 2θ / g

How we know the maximum height

$v_{f}$ ² = $v_{oy}$ ² - 2 g y

$v_{f}$ = 0

$v_{oy}$ = √ 2 g y

$v_{oy}$ = √ 2 9.8 / 15

$v_{oy}$ = 1.14 m / s

Let's use trigonometry to find the speed

sin θ = $v_{oy}$ / vo

vo = $v_{oy}$ / sin θ

vo = 1.14 / sin 60

vo = 1.32 m / s

We calculate the range with the first equation

R = 1.32² sin(2 60) / 30

R = 0.0503 m

3 0

1 year ago

The electric field near the earth's surface has magnitude of about 150n/c. what is the acceleration experienced by an electron n

qaws [65]

Felectric = q*E
<span> Ftranslational = m*a
</span><span> Felectric = Ftranslational
</span> <span>q*E = m*a
</span><span> Solve for a
</span><span> a = q/m*E </span>
<span> Our sign convention is "up is positive"
</span><span> q = 1.6*10^-19 C
</span><span> m = 1.67*10^-27 kg
</span><span> E = -150 N/C (- because it is down and up is positive)
</span> a =<span> -6,4*10^5</span><span> m/s^2 (downward)
</span> answer
a = -6,4*10^5 m/s^2 (downward)

3 0

1 year ago

Engineers who design battery-operated devices such as cell phones and MP3 players try to make them as efficient as possible. An

Svetradugi [14.3K]

efficiency= [useful energy transferred ÷ total energy supply]×100%

So, [5500÷10000]×100%=0.55×100

=55%

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1 year ago

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

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1 year ago