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Aloiza [94]
2 years ago
7

When a scientific calculator shows the quantity below, what does it mean? 1.5E8

Physics
1 answer:
jenyasd209 [6]2 years ago
6 0
1.5 with 8 exponent of 10.
1.5 x 100000000
150000000
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A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg
alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = -\frac{v}{u}

where

u = object distance

v = image distance

Now,

1.90 = \frac{v}{u}

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}

\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}

\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}

\frac{1}{16.5} = \frac{ 0.90}{1.90u}

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0
3 years ago
Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E
Kay [80]

Answer:

The true course: 40.29^\circ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

  • V_w = velocity of wind = 30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h
  • V_p = velocity of plane in still air = 100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h

Assume:

  • V_r = resultant velocity of the plane
  • \theta = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h

Let us find the direction of this resultant velocity with respect to east direction:

\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ

This means the the true course of the plane is in the direction of 40.29^\circ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h

Hence, the ground speed of the plane is 96.68 km/h.

5 0
3 years ago
If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum dista
andrew11 [14]

Answer:

x=2d

Explanation:

initial stretch in the spring is d

so using Hook's law

at equilibrium position

k×d=mg

where k= spring constant

m= mass of fish

g= acceleration due to gravity.

d=mg/k ................ (1)

in second case  by energy conservation

1/2 kx^2=mgx

x=2mg/k

using equation 1

x=2d

3 0
3 years ago
What is the mass of an object if a force of 34 N produces an acceleration of 4.0 m/s squared​
mrs_skeptik [129]

Answer:

8.5 kg

Explanation:

F = ma

34 N = m(4.0m/s^2)

m = 8.5 kg

6 0
2 years ago
Calculate the distance travel if it accelerates from 0 to 27.8 meters per second in 2.5 seconds
luda_lava [24]

The distance travel is 69.5 meters.

<u>Explanation:</u>

Given datas are as follows

Speed = 27.8 meters / second

Time = 2.5 seconds

The formula to calculate the speed using distance and time is

Speed = Distance ÷ Time (units)

Then Distance = Speed × Time (units)

Distance = (27.8 × 2.5) meters          

Distance = 69.50 meters

Therefore the distance travelled is 69.50 meters.

8 0
3 years ago
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