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Vladimir79 [104]

2 years ago

10

the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bott

emvof the ramp

1 answer:

IRISSAK [1]2 years ago

5 0

A

Explanation:

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Find the uniform acceleration that causes a car's velocity to change from 20.0 m/s to 105 m/s in and 12.0 s

Sauron [17]

Answer:

7.08 m/s²

Explanation:

Given:

v₀ = 20.0 m/s

v = 105 m/s

t = 12.0 s

Find: a

v = at + v₀

105 m/s = a (12.0 s) + 20.0 m/s

a = 7.08 m/s²

5 0

3 years ago

Engineers tasked with building a car bumper need high-quality plastic that is readily available. They found the material polycar

jenyasd209 [6]

Answer:

The material must be durable (quality of the material requirement)

Explanation:

The design criteria set for the materials used for technological design are;

1) The materials should be affordable (less costly)

2) The materials should be last for a long duration (high durability)

3) The material should be readily available (easily sourced)

Therefore, given that the engineers initially had the criteria for the required plastic to be of high quality and to be readily available, and that the poly-carbonate they found is long lasting and not too costly, the criteria met that was set initially was the quality criteria of durability.

7 0

3 years ago

Read 2 more answers

If the mass of a material is 50 grams and the volume of the material is 12 cm^3, what would the density of the material be?

AlekseyPX

The density of the material would be
25/6 grams per cm^3.
to obtain the result above this is what we do:
density is calculated as: (the mass of the given material or object) / volume of the material
which leads us to 50grams /12cm^3

4 0

3 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

Pls help and thank u need asap!

aleksandr82 [10.1K]

Answer:

ill help if u help me?its 4am

5 0

2 years ago