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KatRina [158]
3 years ago
12

Two friends grab different sides of a textbook and pull with forces of 6.7 N to the east and 4.4 N to the west, respectively. Wh

at would be the magnitude and direction of the force a third friend would need to exert on the textbook in order to balance the other two forces?
Physics
1 answer:
elena-s [515]3 years ago
8 0

Explanation:

This equation for acceleration can be used to calculate the acceleration of an object that is acted on by a net force. For example, Xander and his scooter have a total mass of 50 kilograms. Assume that the net force acting on Xander and the scooter is 25 Newtons. What is his acceleration? Substitute the relevant values into the equation for acceleration:

<h2>Answer:</h2>

a=Fm=25 N50 kg=0.5 Nkg

The Newton is the SI unit for force. It is defined as the force needed to cause a 1-kilogram mass to accelerate at 1 m/s2. Therefore, force can also be expressed in the unit kg • m/s2. This way of expressing force can be substituted for Newtons in Xander’s acceleration so the answer is expressed in the SI unit for acceleration, which is m/s2:

a=0.5 Nkg=0.5 kg⋅m/s2kg=0.5 m/s2

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A car starts from rest and accelerates at 2m/s² for 30 seconds. He then travel and constant speed for 1 minute and finally decel
zhenek [66]

Answer:

60

Explanation:

15 miles

6 0
3 years ago
A 1.0 × 10^3 kg sports car is initially traveling at 15 m/s. The driver then applies the brakes for several seconds so that -25
algol13

Answer:

Ea = 112500[J]

Eb = 87500[J]

Explanation:

To solve this problem we must use the principle of energy conservation which tells us that the energy of a body plus the work done or applied by the body equals the final energy of a body.

This can be easily visualized by the following equation:

E_{A}+E_{friction}=E_{B}

Now we must define the energies at points A & B.

<u>For point A</u>

At point A we only have kinetic energy since it moves at 15 [m/s]

So the kinetic energy

E_{A}=\frac{1}{2}*m*v_{A}^{2}  \\E_{A}=\frac{1}{2} *1000*(15)^{2} \\E_{A}=112500[J]

The final kinetic energy can be calculated as follows:

112500-25000=E_{B}\\E_{B}=87500[J]

8 0
3 years ago
A 62.2-kg person, running horizontally with a velocity of 3.80 m/s, jumps onto a 19.7-kg sled that is initially at rest. (a) Ign
jek_recluse [69]

Answer:

a) v = 2.886\,\frac{m}{s}, b) \mu_{k} = 0.014

Explanation:

a) The final speed is determined by the Principle of Momentum Conservation:

(62.2\,kg)\cdot (3.80\,\frac{m}{s} ) = (81.9\,kg)\cdot v

v = 2.886\,\frac{m}{s}

b) The deceleration experimented by the system person-sled is:

a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(2.886\,\frac{m}{s} \right)^{2}}{2\cdot (30\,m)}

a = -0.139\,\frac{m}{s^{2}}

By using the Newton's Laws, the only force acting on the motion of the system is the friction between snow and sled. The kinetic coefficient of friction is:

-\mu_{k}\cdot m\cdot g = m\cdot a

\mu_{k} = -\frac{a}{g}

\mu_{k} = -\frac{\left(-0.139\,\frac{m}{s^{2}} \right)}{9.807\,\frac{m}{s^{2}} }

\mu_{k} = 0.014

3 0
4 years ago
All of the forces that act on an object add up to equal________________
sveta [45]
Net force I’m pretty sure
4 0
4 years ago
A proton is first accelerated from rest through a potential difference V and then enters a uniform 0.750-T magnetic field orient
galben [10]

The magnetic force acting on a charged particle moving perpendicular to the field is:

F_{b} = qvB

F_{b} is the magnetic force, q is the particle charge, v is the particle velocity, and B is the magnetic field strength.

The centripetal force acting on a particle moving in a circular path is:

F_{c} = mv²/r

F_{c} is the centripetal force, m is the mass, v is the particle velocity, and r is the radius of the circular path.

If the magnetic force is acting as the centripetal force, set F_{b} equal to F_{c} and solve for v:

qvB = mv²/r

v = qBr/m

Due to the work-energy theorem, the work done on the proton by the potential difference V becomes the proton's kinetic energy:

W = KE

W is work, KE is kinetic energy

W = Vq

KE = 0.5mv²

Therefore:

Vq = 0.5mv²

Substitute v = qBr/m and solve for V:

V = 0.5qB²r²/m

Given values:

m = 1.67×10⁻²⁷kg (proton mass)

B = 0.750T

q = 1.60×10⁻¹⁹C (proton charge)

r = 1.84×10⁻²m

Plug in the values and solve for V:

V = (0.5)(1.60×10⁻¹⁹)(0.750)²(1.84×10⁻²)²/1.67×10⁻²⁷

V = 9120V

6 0
4 years ago
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