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Furkat [3]
3 years ago
13

What is the weight of a ring tailed lemur that has a mass of 10 kg? Use w = mg (g = -9.8 m/s^2)

Physics
1 answer:
marusya05 [52]3 years ago
4 0

m = 10 kg

g = -9.8 m/s2

w = m * g

w= -9.8 * 10

w= -98 N

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A main goal of most environmental scientists is to achieve _________________________.
sergiy2304 [10]

Answer:

The answer is biodiversity

Explanation:

4 0
3 years ago
A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th
Norma-Jean [14]

Answer:

0.29 m

Explanation:

9 mm = 0.009 m in diameter

Cross-sectional area A = \pi d^2/4 = \pi * 0.009^2/4 = 6.36\times 10^{-5} m^2

Let the tensile modulus of Nickel E = 170 \times 10^9Pa.

The elongation of the rod can be calculated using the following formula:

\Delta L = \frac{F L}{A E} = \frac{6283*50}{6.36\times 10^{-5} * 170 \times 10^9} = \frac{314150}{1081200} = 0.29 m

6 0
3 years ago
If anyone here can help me with physics please comment. its about coulombs law
Minchanka [31]
I may be able to help you with This subject
3 0
3 years ago
A net force of 3000 N is accelerating a 1200 kg elevator upward. If the elevator starts from rest, how long will it take to trav
alexdok [17]

F = net force acting on the elevator in upward direction = 3000 N

m = mass of the elevator = 1200 kg

a = acceleration of the elevator = ?

Acceleration of the elevator is given as

a = F/m

a = 3000/1200

a = 2.5 m/s²

v₀ = initial velocity of the elevator = 0 m/s

Y = displacement of the elevator = 15 m

t = time taken

Using the kinematics equation

Y = v₀ t + (0.5) a t²

15 = (0) t + (0.5) (2.5) t²

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5 0
3 years ago
When tightening a bolt, you push perpendicularly on a wrench with a force of 165 N at a distance of 0.140 m from the center of t
choli [55]

Answer:

Part a)

\tau = 23.1 Nm

Part b)

\tau = 17.05 Foot pound force

Explanation:

As we know that torque is defined as the product of force and its perpendicular distance from reference point

so here we have

\tau = \vec r \times \vec F

now we have

\tau = (0.140)(165)

\tau = 23.1 Nm

Part b)

Now we know the conversion as

1 meter = 3.28 foot

1 N = 0.225 Lb force

now we have

\tau = 23.1 Nm

\tau = 23.1 (0.225 Lb)(3.28 foot)

\tau = 17.05 Foot pound force

3 0
3 years ago
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