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Elan Coil [88]
3 years ago
13

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

ds?
Physics
1 answer:
taurus [48]3 years ago
4 0

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

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Please help !! In the diagram, q1 = +0.00200 C, q2 = 0.00180 C, and q3 = +0.00830 C. the net force on q2 is zero. how far is q2
VikaD [51]

Answer:

2.03715

Explanation:

32364=8.99\cdot 10^9\cdot \frac{0.00180\cdot 0.00830}{2.03715^2}

4 0
3 years ago
_______ are different forms of a single element. A) Atoms B) Elements C) Ions D) Isotopes
ELEN [110]
The answer is D. Isotopes.
Hope that helped.

7 0
4 years ago
Read 2 more answers
g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin
DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as mgh=0.5mv^{2} since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then v=\sqrt {2gh} and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s

6 0
3 years ago
A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil
snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m

The locations are +5.9 cm or -5.9 cm

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3 years ago
How does the observed change in the sea floor age support he theory of sea floor spreading
RUDIKE [14]
Dna is the way to answer this questions
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