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stich3 [128]
3 years ago
6

sing a rope that will snap if the tension in it exceeds 361 N, you need to lower a bundle of old roofing material weighing 455 N

from a point 6.80 m above the ground. (a) What magnitude of the bundle's acceleration will put the rope on the verge of snapping? (b) At that acceleration, with what speed would the bundle hit the ground

Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

a)-2m/s^2

b)27.2m/s

Explanation:

Hello! The first step to solve this problem is to find the mass of the block remembering that the definition of weight force is mass by gravity (g=9.8m / s ^ 2)

W=455N=weight

W=mg

W=455N=weight

m=\frac{W}{g} =\frac{455}{9.81}=46.38kg

The second step is to draw the free body diagram of the body (see attached image) and use Newton's second law that states that the sum of the forces is equal to mass by acceleration

a=\frac{T-W}{m} =\frac{361-455}{46.38kg} =-2m/s^2

for point b we use the equations of motion with constant acceleration to find the velocity

Vf=\sqrt{X(2)(a)+Vo^2}

Where

Vf = final speed

Vo = Initial speed =0

A = acceleration =2m/s

X = displacement =6.8m

Solving

Vf=\sqrt{(6.8)(2)(2)+0^2}=27.2m/s

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Select the correct answer.
Dahasolnce [82]

Answer:

a is the correct choice

Explanation:

5 0
3 years ago
Read 2 more answers
A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2 , and 1.85 ms (1 ms = 10−3
ankoles [38]

Answer:

u = - 38.85 m/s^-1

Explanation:

given data:

acceleration = 2.10*10^4 m/s^2

time = 1.85*10^{-3} s

final velocity = 0 m/s

from equation of motion we have following relation

v = u +at

0 =  u + 2.10*10^4 *1.85*10^{-3}

0 = u + (21 *1.85)

0 = u + 38.85

u = - 38.85 m/s^-1

negative sign indicate that the ball bounce in opposite directon

4 0
3 years ago
A 90 kg ice skater moving at 12.0 m/s on the ice encounters a region of roughed up ice with a coefficient of kinetic friction of
balandron [24]

Answer:

The skater covers a distance of <u>15 m</u> before stopping.

Explanation:

Let the distance traveled before stopping be 'd' m.

Given:

Mass of the skater (m) = 90 kg

Initial velocity of the skater (u) = 12.0 m/s

Final velocity of the skater (v) = 0 m/s (Stops finally)

Coefficient of kinetic friction (μ) = 0.490

Acceleration due to gravity (g) = 9.8 m/s²

Now, we know that, from work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.

Here, the net force acting on the skater is only frictional force which acts in the direction opposite to motion.

Frictional force is given as:

f=\mu N

Where, 'N' is the normal force acting on the skater. As there is no vertical motion, N=mg

∴ f=\mu mg=0.490\times 90\times 9.8=432.18\ N

Now, work done by friction is a negative work as friction and displacement are in opposite direction and is given as:

W=-fd=-432.18d

Now, change in kinetic energy is given as:

\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}\times 90(0-12^2)\\\\\Delta K=45\times (-144)=-6480\ J

Therefore, from work-energy theorem,

W=\Delta K\\\\-432.18d=6480\\\\d=\frac{6480}{432.18}\\\\d=14.99\approx 15\ m

Hence, the skater covers a distance of 15 m before stopping.

7 0
3 years ago
List 3 examples of transition metals
STatiana [176]

Answer:

Explanation:

Titanium

Chromium.

Iron

4 0
3 years ago
Name of a body that changes Chemical energy to electric energy.​
Ratling [72]

Answer:

battery

Explanation:

A battery contains stored chemical energy and converts it to electrical energy. (cK.12)

6 0
3 years ago
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