Ask question

Ask question

All categories

Katyanochek1 [597]

3 years ago

15

A student travels 4.0m East, 8.0m South, 4.0m west, and finally 8.0m north. What is her displacement? What distance did she trav

el?

1 answer:

erica [24]3 years ago

3 0

no cap bro bro i thinks its poop Explanation:

You might be interested in

Urgent! please help!

kap26 [50]

$_{94}^{239}Pu+ _{0}^{1}n--->_{40}^{100}Zr+_{x}^{y}Element+2_{0}^{1}n\\ \\ 94+0=40+x+0, x= 54,\\ \\ 239+1 = 100+y+2*1, y=138\\ \\_{54}^{138}Element= _{54}^{138}Xe$

Answer is A.

8 0

3 years ago

Calculate the size of the image of a tree that is 8m high and 80 m a pinhole camera that is 20 cm long . what is its magnificati

Vladimir79 [104]

1) Size of the image: 2 cm

In order to calculate the size of the image, we can use the following proportion:

$p:q = h_o : h_i$

where

p = 80 m is the distance of the tree from the pinhole

q = 20 cm = 0.2 m is the distance of the image from the pinhole

$h_o$ = 8 m is the heigth of the object

$h_i$ is the height of the image

By re-arranging the proportion, we find

$h_i = \frac{h_o \cdot q}{p}=\frac{(8 m)(0.2 m)}{80 m}=0.02 m=2 cm$

2) Magnification: 0.0025

The magnification of a camera is given by the ratio between the size of the image and the size of the real object:

$M=\frac{h_i}{h_o}$

so, in this problem we have

$M=\frac{0.02 m}{8 m}=0.0025$

4 0

3 years ago

A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its

bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration $\alpha =140rad/sec^2$

We know that torque $\tau =I\alpha$

And $\tau =Fr$

So $Fr=I\alpha$ , here I is moment of inertia

So $50\times 0.12=I\times 140$

$I=0.0428kgm^2$

We know that moment of inertia $I=\frac{1}{2}mr^2$

So $0.0428=\frac{1}{2}\times m\times 0.12^2$

m = 2.976 kg

6 0

3 years ago

An electron is initially moving at 1.4 x 107 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude 120 N

algol13

Answer:

K.E = 15.57 x 10⁻¹⁷ J

Explanation:

First, we find the acceleration of the electron by using the formula of electric field:

E = F/q

F = Eq

but, from Newton's 2nd Law:

F = ma

Comparing both equations, we get:

ma = Eq

a = Eq/m

where,

E = electric field intensity = 120 N/C

q = charge of electron = 1.6 x 10⁻¹⁹ C

m = Mass of electron = 9.1 x 10⁻³¹ kg

Therefore,

a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)

a = 2.11 x 10¹³ m/s²

Now, we need to find the final velocity of the electron. Using 3rd equation of motion:

2as = Vf² - Vi²

where,

Vf = Final Velocity = ?

Vi = Initial Velocity = 1.4 x 10⁷ m/s

s = distance = 3.5 m

Therefore,

(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²

Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)

Vf = 1.85 x 10⁷ m/s

Now, we find the kinetic energy of electron at the end of the motion:

K.E = (0.5)(m)(Vf)²

K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²

<u>K.E = 15.57 x 10⁻¹⁷ J</u>

4 0

3 years ago

Consider a 2 m wide and 5 m long slab resting on flat, fixed, earthen bed. The slab is 20 cm thick with uniform properties. You

4vir4ik [10]

Answer:

Explanation:

The coordinate sketch for the system is shown in the attached file below. Also, in the cartesian coordinate system, since the height is less than the length and width, we did neglect the height. Thus, we eliminate the height and converted it to a two-dimension.

8 0

3 years ago