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ira [324]
3 years ago
10

Which phrase best completes the diagram?

Physics
2 answers:
xenn [34]3 years ago
7 0

Answer:

D. makes rules for other banks

Explanation:

FromTheMoon [43]3 years ago
5 0

Answer:

B

Explanation:

i think

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3.
Vladimir [108]

Answer:

C. Resist change in motion.

Explanation:

Inertia: the tendency of an object to resist a change in motion unless an outside force acts on the object.

8 0
3 years ago
Read 2 more answers
Jacques is trying to put a screw into a hard piece of wood, but it is too difficult for him to turn. Which of the following woul
solniwko [45]

Answer:

A. It is the only one that makes sense

Explanation:

3 0
4 years ago
An object of mass m is initially at rest and free to move without friction in any direction in the xy-plane. A constant net forc
Dominik [7]

Answer:3/5

Explanation:

6/6

3 0
3 years ago
2. A 15 kg mass fastened to the end of a steel wire of un-
Flauer [41]

Explanation:

Elongation of the wire is:

ΔL = F L₀ / (E A)

where F is the force,

L₀ is the initial length,

E is Young's modulus,

and A is the cross sectional area.

ΔL = T (0.5 m) / ((2.0×10¹¹ Pa) (0.02 cm²) (1 m / 100 cm)²)

ΔL = T (1.25×10⁻⁶ m/N)

T = (80,000 N/m) ΔL

Draw a free body diagram of the mass at the bottom of the circle.  There are two forces: tension force T pulling up and weight force mg pulling down.

Sum of forces in the centripetal direction:

∑F = ma

T − mg = mv²/r

T − mg = mω²r

T − (15 kg) (9.8 m/s²) = (15 kg) (2 rev/s × 2π rad/rev)² (0.5 m + ΔL)

T − 147 N = (2368.7 N/m) (0.5 m + ΔL)

Substitute:

(80,000 N/m) ΔL − 147 N = (2368.7 N/m) (0.5 m + ΔL)

(80,000 N/m) ΔL − 147 N = 1184.35 N + (2368.7 N/m) ΔL

(797631.3 N/m) ΔL = 1331.35 N

ΔL = 0.00167 m

ΔL = 1.67 mm

6 0
3 years ago
A 100-kg running back runs at 5 m/s into a stationary linebacker. It takes 0.5 s for the running back to be completely stopped.
Elza [17]

Answer:

1000 N

Explanation:

First, we need to find the deceleration of the running back, which is given by:

a=\frac{v-u}{t}

where

v = 0 is his final velocity

u = 5 m/s is his initial velocity

t = 0.5 s is the time taken

Substituting, we have

a=\frac{0-5 m/s}{0.5 s}=-10 m/s^2

And now we can calculate the force exerted on the running back, by using Newton's second law:

F=ma=(100 kg)(-10 m/s^2)=-1000 N

so, the magnitude of the force is 1000 N.

6 0
4 years ago
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