What is the mass of 5.4 moles of copper ?

What does the Bohr Model show us?

aivan3 [116]

Answer:

2

Explanation:

5 0

3 years ago

What is the maximum force that could be applied to anterior cruciate ligament (ACL) if it has a diameter of 4.8 mm and a tensile

vovangra [49]

Answer:

Maximum force, F = 1809.55 N

Explanation:

Given that,

Diameter of the anterior cruciate ligament, d = 4.8 mm

Radius, r = 2.4 mm

The tensile strength of the anterior cruciate ligament, $P=100\times 10^6\ N/m^2=10^8\ Pa$

We need to find the maximum force that could be applied to anterior cruciate ligament. We know that the unit of tensile strength is Pa. It must be a type of pressure. So,

$F=P\times A\\\\F=10^8\times \pi (2.4\times 10^{-3})^2\\\\F=1809.55\ N$

So, the maximum force that could be applied to anterior cruciate ligament is 1809.55 N

4 0

4 years ago

A spring with spring constant 15.5 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

Aloiza [94]

Answer:

A) 138.8g

B)73.97 cm/s

Explanation:

K = 15.5 Kn/m

A = 7 cm

N = 37 oscillations

tn = 20 seconds

A) In harmonic motion, we know that;

ω² = k/m and m = k/ω²

Also, angular frequency (ω) = 2π/T

Now, T is the time it takes to complete one oscillation.

So from the question, we can calculate T as;

T = 22/37.

Thus ;

ω = 2π/(22/37) = 10.5672

So,mass of ball (m) = k/ω² = 15.5/10.5672² = 0.1388kg or 138.8g

B) In simple harmonic motion, velocity is given as;

v(t) = vmax Sin (ωt + Φ)

It is from the derivative of;

v(t) = -Aω Sin (ωt + Φ)

So comparing the two equations of v(t), we can see that ;

vmax = Aω

Vmax = 7 x 10.5672 = 73.97 cm/s

6 0

3 years ago

A Porsche challenges a Honda to a 200-m race.Because the Porsche's acceleration of 3.5 m/s2 is larger than the Honda's 3.0 m/s2,

Blizzard [7]

Answer:

Honda won by 0.14 s

Explanation:

We are given that

Distance =S=200 m

Initial velocity of Honda=u=0m/s

Initial velocity of Porsche=u'=0m/s

Acceleration of Honda= $3.0m/s^2$

Acceleration of Porsche's= $3.5m/s^2$

Time taken by Honda to start=1 s

$s=ut+\frac{1}{2}at^2$

Substitute the values

$200=0(t)+\frac{1}{2}(3)t^2$

$200=\frac{3}{2}t^2$

$t^2=\frac{200\times 2}{3}=\frac{400}{3}$

$t=\sqrt{\frac{400}{3}}=11.55s$

Time taken by Honda=11.55 s

Now, time taken by Porsche

$200=\frac{1}{2}(3.5)t^2$

$t^2=\frac{200\times 2}{3.5}$

$t=\sqrt{\frac{400}{3.5}}=10.69 s$

Total time taken by Porsche=10.69+1=11.69 s

Because it start 1 s late

Time taken by Honda is less than Porsche .Therefore, Honda won and

Time =11.69-11.55=0.14 s

Honda won by 0.14 s

3 0

3 years ago

A 56.88 kg diver hits the water going 10 m/s. From what height in meters platform did she jump?

Otrada [13]

Answer:

its 10

Explanation:

if you search it up you will see

8 0

3 years ago