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3 years ago

14

i really need help! please hurry and answer this. i'll give you 5 stars and everything just please answer this as fast as you ca

n.

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer:

Triple, double, single

Explanation:

They are all the same type of elemental bond, therefore the more of the same bond, you can only assume it would get stronger.

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What is the scientific definition of tracked?

dexar [7]

To find or discover by investigation?

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3 years ago

Bats are extremely adept at catching insects in midair. If a 81.5-g bat flying in one direction at 7.21 m/s catches a 8.11-g ins

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3 years ago

The 1 kg box is sliding along a frictionless surface. It collides with and sticks to the 2 kg box. Afterward, the speed of the t

Nina [5.8K]

Answer:

The speed of the boxes are 1 m/s.

Explanation:

Given that,

Mass of box = 1 kg

Mass of another box = 2 kg

Suppose 1 kg box moves with 3 m/s speed.

We need to calculate the speed of the boxes

Using formula of conservation of momentum

$m_{1}u_{1}+m_{2}u_{2}=(m_{1}+m_{2})v$

Where, u = initial velocity

v = final velocity

Put the value into the formula

$1\times3+2\times0=(1+2)v$

$v=\dfrac{3}{3}$

$v=1\ m/s$

Hence, The speed of the boxes are 1 m/s.

4 0

3 years ago

A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio

Andru [333]

To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

$f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}$

$f=2.85 Hz$

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

$\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2$

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

$kA^2 =mV^2 +ky^2$

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

$1700*A^2=5.3*1.7^2 +1700*(0.045)^2$

Solving for A,

$A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}$

$A ^2 = 0.011035$

$A=0.105 m \approx 10.5 cm$

PART C) Finally, the total mechanical energy is given by the equation

$E = \frac{1}{2}kA^2$

$E=\frac{1}{2}1700*(0.105)^2$

$E= 9.3712 J$

3 0

3 years ago

A sled slides along a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.6 m/

jenyasd209 [6]

Answer:

$\Delta t =1.31\ s$

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

$J = m(v_f - v_i)$

frictional force

F = μ N

where as N is the normal force

now,

$F\Delta t = m(v_f -v_i)$

$\mu m g \times \Delta t = m(v_f-v_i)$

$\mu g \times \Delta t = v_f-v_i$

$\Delta t =\dfrac{v_f-v_i}{\mu g}$

$\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}$

$\Delta t =1.31\ s$

time taken to move from A to B is equal to 1.31 s

3 0

3 years ago

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