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never [62]
4 years ago
11

a 1.5-kilogram cart initially moves at 2.0 meters per second. it is brought to rest by a constant net force in 0.30 second. what

is the magnitude of the net force
Physics
1 answer:
KiRa [710]4 years ago
7 0
Since acceleration is constant, we can use kinematic equation v=u+at( v= final velocity = 0...since finally cart stops. u = initial velocity = 2m/s, a = deacceleration and t= 0.3 sec). Therefore 0 = 2 - a(0.3). a = 20/3 = 6.67m/s^2. Net force = massxacceleration = 1.5x6.67 = 10N. 
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The cheetah, the fastest of land animals, can run a distance of 274 m in 8.65 s at its top speed. What is the cheetahs top speed
OverLord2011 [107]

Answer:

31.68 meters per second.

Explanation:

Speed is equals to Distance divided by Time, so, this means that speed is inversely proportional to time and directly proportional to distance.

Here, Distance is given which is 274 meters.

          Time taken to cover the given distance is 8.65 seconds.

So, by putting the values of distance and time in the formula of speed we will get the top speed of Cheetah which is 31.68 meters per second.

8 0
3 years ago
3. A 0.35 kg puck slides across the ice with an average force of friction of 0.15 N acting on it. It slides 82 m before coming t
sattari [20]

Answer:

Work done is 12.3 J

Explanation:

We have,

Mass of puck, m = 0.35 kg

Force of friction acting on the puck when it slides is 0.15 N

Distance travelled by the puck is 82 m.

It is required to find the work done on the puck. Finally the puck comes to rest and the force of friction is acting on it. It means the applied force is 0.15 N. Work done is given by

W=Fd\\\\W=0.15\times 82\\\\W=12.3\ J

The work done on the puck is 12.3 J.

5 0
4 years ago
How many cells must be connected in series to give the 350 v a large catfish can produce?
victus00 [196]
Each electrocyte can produce 110 mv.
110 m/V=0.110V
n(0.110)=350
n=3182 (rounded)
3 0
3 years ago
A 90 kg person stands at the edge of a stationary children's merry-go-round at a distance of 5.0 m from its center. The person s
Paraphin [41]

Answer:

\omega = 0.016\,\frac{rad}{s}

Explanation:

The rotation rate of the man is:

\omega = \frac{v}{R}

\omega = \frac{0.80\,\frac{m}{s} }{5\,m}

\omega = 0.16\,\frac{rad}{s}

The resultant rotation rate of the system is computed from the Principle of Angular Momentum Conservation:

(90\,kg)\cdot (5\,m)^{2}\cdot (0.16\,\frac{rad}{s} ) = [(90\,kg)\cdot (5\,m)^{2}+20000\,kg\cdot m^{2}]\cdot \omega

The final angular speed is:

\omega = 0.016\,\frac{rad}{s}

3 0
3 years ago
A circular test track for cars has a circumference of 3.5 km . A car travels around the track from the southernmost point to the
Marta_Voda [28]
<span>The car would have traveled exactly one-half of the circumference of the track, since it would have gone from one extreme point to its opposite extreme point. This would be equal to (3.5 / 2), or 1.75 km. The northernmost point would be 1.75km away from the southernmost point.</span>
8 0
4 years ago
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