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stealth61 [152]
2 years ago
8

You want to examine the hairy details of your favorite pet caterpillar, using a lens of focal length 8.9 cm 8.9 cm that you just

happen to have around. With the lens close to your eye and the animal at the lens's focal point, what angular magnification M M do you achieve? Assume your near point is at 25.0 cm.
Physics
1 answer:
Zepler [3.9K]2 years ago
4 0

Answer:

The angular magnification is M = 2.808

Explanation:

From the question we are told

           The focal length is  f = 8.9cm

          The near point is n_p = 25.0cm

The angular magnification is mathematically represented as

                          M = \frac{n_p}{f}

Substituting values

                        M = \frac{25}{8.9}

                           = 2.808

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A person driving a car suddenly applies the brakes. The car takes 4 s to come to rest while traveling 20 m at constant accelerat
Zinaida [17]

Answer:

Yes we can find the initial velocity of car without finding acceleration.

u = 10 m/s.

Explanation:

Given that

s=20 m

Car takes 4 s to come in rest.

We know that when acceleration is constant then we can apply motion equation

v=u+at        ----------1

s=ut+\dfrac{1}{2}at^2       ------2

From equation 1 and 2

s=ut+\dfrac{1}{2}t^2\left (\dfrac{v-u}{t} \right )

So we can say that

s= \left(\dfrac{v+u}{2}\right)t

Given that the velocity of car at final condition will be zero (v=0)

s= \left(\dfrac{0+u}{2}\right)t

s= \left(\dfrac{u}{2}\right)t

From the above equation we can find the initial velocity of car without finding the acceleration

20= \dfrac{u}{2}\times 4

u = 10 m/s

7 0
2 years ago
A vertical spring of constant k = 400 N/m hangs at rest. When a 2 kg mass is attached to it, and it is released, the spring exte
Viefleur [7K]

Answer:

4.9 cm

Explanation:

From Hook's Law,

F = ke......................... Equation 1

Where F= force, e = extension, k = spring constant.

Note: the Force acting on the the spring is the weight of the mass.

W = mg.

F = mg.................... Equation 2

Where m = mass, g = acceleration due to gravity

Substitute equation 2 into equation 1

mg = ke

make e the subject of the equation

e = mg/k............... Equation 3.

Given: m = 2 kg, g = 9.8 m/s², k = 400 N/m

e = (2×9.8)/400

e = 19.6/400

e = 0.049 m

e = 4.9 cm

3 0
2 years ago
A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin
krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

F=(0.850) sin (\frac{x}{2.00}) [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

a=\frac{F}{m}

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00}) [m/s^2]

When x = 3.14 m, the acceleration is:

a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2

Now we can find the final speed of the particle by using the suvat equation:

v^2-u^2=2ax

where

u = 8.00 m/s is the initial velocity

v is the final velocity

a=5.67 m/s^2

x = 3.14 m is the displacement

Solving for v,

v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s

And the speed is just the magnitude of the velocity, so 9.98 m/s.

4 0
2 years ago
Grace, Erin, and Tony are on a seesaw. Grace has a mass of 45kg and is seated 0.7m to the left of the fulcrum. Nicole has a mass
salantis [7]
Use Scoratic it works with any time of subject
5 0
2 years ago
A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of ar
Elena L [17]

Answer:

assume nitrogen is an ideal gas with cv=5R/2

assume argon is an ideal gas with cv=3R/2

n1=4moles

n2=2.5 moles

t1=75°C   <em>in kelvin</em> t1=75+273

t1=348K

T2=130°C  <em>in kelvin</em> t2=130+273

t2=403K

u=пCVΔT

U(N₂)+U(Argon)=0

<em>putting values:</em>

=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)

<em>by simplifying:</em>

Tfinal=363K

6 0
2 years ago
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