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faust18 [17]
3 years ago
9

I need help, please!

Physics
1 answer:
marta [7]3 years ago
7 0

NO. it is not balanced

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A 26.3 kg object is traveling at 21.0 m/s north. What average net force is required to bring this object to a stop in 2.60 secon
taurus [48]

The average force is -212.4 N

Explanation:

We can solve this problem by using the impulse theorem, which states that the impulse exerted on the object (the product of the force exerted and the time) is equal to the change in momentum of the object:

F \Delta t = m \Delta v

where

F is the net force on the object

\Delta t is the time

m is the mass

\Delta v is the change in velocity

In this problem, we have:

m = 26.3 kg

\Delta v = -21.0 m/s

\Delta t = 2.60 s

Solving for F, we find

F=\frac{m\Delta v}{\Delta t}=\frac{(26.3)(-21.0)}{2.60}=-212.4 N

where the negative sign indicates that the direction of the force is opposite to the motion of the object.

Learn more about force and momentum:

brainly.com/question/9484203

#LearnwithBrainly

4 0
3 years ago
A construction worker puts 20J of energy in to one strike of his hammer on the head of a nail. The energy transferred to driving
Slav-nsk [51]

Answer:

efficiancy=40 percent

Explanation:

efficiency=energy output/energy input×100

efficiancy=8J/20J×100

efficiancy=0.4×100

efficiancy=40 percent

Mark brianliest if my answer suit your question..

3 0
3 years ago
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A 460 g , 6.0-cm-diameter can is filled with uniform, dense food. It rolls across the floor at 1.1 m/s . Part A What is the can'
Reika [66]

Answer:

the can's kinetic energy is 0.42 J

Explanation:

given information:

Mass, m = 460 g = 0.46 kg

diameter, d = 6 cm, so r = d/2 = 6/2 = 3 cm = 0.03 m

velocity, v = 1.1 m/s

the kinetic energy of the can is the total of kinetic energy of the translation and rotational.

KE = \frac{1}{2} I ω^2 + \frac{1}{2} mv^{2}

where

I = \frac{1}{2} mr^{2} and ω = \frac{v}{r}

thus,

KE = \frac{1}{2} \frac{1}{2} mr^{2} (\frac{v}{r})^2 + \frac{1}{2} mv^{2}

     = \frac{1}{2} \frac{1}{2} mr^{2} \frac{v^{2} }{r^{2}} + \frac{1}{2} mv^{2}

     = \frac{1}{4} mv^{2} + \frac{1}{2} mv^{2}

     = \frac{3}{4} mv^{2}

     = \frac{3}{4} (0.46) (1.1)^{2}

     = 0.42 J

8 0
3 years ago
How high would a projectile go if it was launched from ground level with an initial speed of 26 m/s at an angle of 30 degrees ab
tigry1 [53]

Answer:

Vy = 26 m/s sin 30 = 13 m/s      vertical speed

t = Vy / a = 13 m/s / 9.80 m/s^2 = 1.33 sec     time to reach Vy = 0

H = Vy t + 1/2 g t^2

H = 13 m/s * 1.33 sec - 1.33^2 * 9.8 / 2 m = 8.62 m

4 0
1 year ago
Which of the following is not a characteristic of a creative person
Katarina [22]
<span>B.Extrinsic motivation </span>
7 0
3 years ago
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