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VARVARA [1.3K]
3 years ago
12

Forces of 11.9 N north, 19.1 N east, and 14.4 N south are simultaneously applied to a 3.77 kg mass as it rests on a frictionless

air table. What is the magnitude of its acceleration?
Physics
1 answer:
oksian1 [2.3K]3 years ago
4 0
Force Vector = <19.1i+11.9j-14.4j>N
Force Vector = <19.1i-2.5j>N
Acceleration Vector =(Force Vector)/mass
Acceleration Vector =<19.1i-2.5j>N/3.77
Acceleration Vector =<5.0663i-0.6631j>m/s^2
|a| = (5.0663^2+(-0.6631)^2)^0.5
|a| = 5.11 (m/s^2)
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Allison wants to determine the density of a bouncing ball. which metric measurements must she use?
lubasha [3.4K]
Density depends on mass and volume so option D is correct answer. Hope this helps!
3 0
3 years ago
Please help need this asap
NikAS [45]

Answer:

A)  350 N

B)  58.33 N

C)  35 kg

D)  35 kg

Explanation:

If we use that g = 10 m/s^2, then the acceleration of gravity on the Moon will be 10/6 m/s^2 = 5/3 m/s*2

The weight of the object on Earth is given by:

Weight = mass * g = 35 * 10 = 350 N

The weight of the object on the Moon:

Weight = mass * gmoon = 35 * 5/3 = 58.33 N

The mass of the object on Earth is 35 kg

The mass of the object on the Moon is exactly the same as on the Earth (35 kg) since the mass is a quantity inherent to the object and not to its location.

3 0
2 years ago
5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co
Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop.  At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0
2 years ago
A ball that has a mass of 0.25 kg spins in a circle at the end of a 1.6 m rope. the ball moves at a tangential speed of 12.2 m/s
NARA [144]

The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.

<h3>What is centripetal force?</h3>

The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.

The given data in the problem is;

m is the mass of A ball = 0.25 kg

r is the radius of circle= 1.6 m rope

v is the tangential speed = 12.2 m/s

\rm F_C is the centripetal force acting on the ball

The centripetal force is found as;

\rm F_C = \frac{mv^2}{r}  \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6}  \\\\ F_C=23.26\ N

Hence the centripetal force acting on the ball will be 23.26 N.

To learn more about the centripetal force refer to the link;

brainly.com/question/10596517

4 0
2 years ago
Is Mercury a heavier element than tin
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Yes because mercury has more protons and electrons that tin. (30 more)
4 0
3 years ago
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