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zysi [14]
2 years ago
6

Pulling on a spring with a force of 1.2 N causes a stretch of 6.4 cm. What is the spring constant for this spring?

Physics
2 answers:
densk [106]2 years ago
7 0

Answer:

k=19 N/m

Explanation:

Use the equation F=kx for this problem and isolate k since that is what you're solving for (k=F/x). Plug in your values (F=1.2 N and x=0.064 m) and solve for k which is 19 N/m.

Genrish500 [490]2 years ago
5 0

Answer:0.1875

Explanation:

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What is the speed of a truck that travels 10 i’m in 10 minutes ?
fiasKO [112]

10km/10min is a legitimate speed. So is meters/sec, km/hour (kph), etc.  

Kph is very common for vehicles:  

10 km/10 min (60 min/hr) = 60 kph

7 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Alja [10]

Answer:

16294 rad/s

Explanation:

Given that

M(ns) = 2M(s), where

M(s) = 1.99*10^30 kg, so that

M(ns) = 3.98*10^30 kg

Again, R(ns) = 10 km

Using the law of gravitation, the force between the Neutron star and the sun is..

F = G.M(ns).M(s) / R²(ns), where

G = 6.67*10^-11, gravitational constant

Again, centripetal force of the neutron star is given as

F = M(ns).v² / R(ns)

Recall that v = wR(ns), so that

F = M(s).w².R(ns)

For a circular motion, it's been established that the centripetal force is equal to the gravitational force, hence

F = F

G.M(ns).M(s) / R²(ns) = M(s).w².R(ns)

Making W subject of formula, we have

w = √[{G.M(ns).M(s) / R²(ns)} / {M(s).R(ns)}]

w = √[{G.M(ns)} / {R³(ns)}]

w = √[(6.67*10^-11 * 3.98*10^30) / 10000³]

w = √[2.655*10^20 / 1*10^12]

w = √(2.655*10^8)

w = 16294 rad/s

7 0
3 years ago
Because white dwarfs are small, as their name implies, they are hard to see. What is a way astronomers have to find white dwarfs
Bumek [7]

Astronomers find white dwarfs that distinguish them from main sequence stars because white dwarfs get really hot, we can search for their ultraviolet radiation.

<h3>What is a white dwarf?</h3>

A white dwarf is a very hot star that radiated much energy in the form of ultraviolet radiation.

This UV radiation is initially very bright and then these stars become red with time.

In conclusion, Astronomers find white dwarfs they can search for their ultraviolet radiation.

Learn more about white dwarfs here:

brainly.com/question/19602278

#SPJ1

3 0
2 years ago
Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra
Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

s^2 = (x_A - x_B)^2+(y_A - y_B)^2  (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

s^2 = x_B^2+y_A^2 (2)

Taking the differential with respect to time:

\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}  (3)

where \displaystyle{\frac{dx_B}{dt}}=v_B and \displaystyle{\frac{dx_A}{dt}}=v_A are the respective given velocities of the boats. To find s and x_B we make use of the given position for A, y_A=30, the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h

with this time, we know can now calculate the distance at which B is:

\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi

and applying Pythagoras:

\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}

Now substituting all the values in (3) and solving for  \displaystyle{\frac{ds}{dt} } we get:

\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}

4 0
3 years ago
A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at
stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft

Also, at any instant t

\Rightarrow l^2=x^2+y^2

differentiate w.r.t.

\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s

5 0
3 years ago
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