All categories

3 years ago

Need help with 6 , 7 , and 8.

Physics

1 answer:

tankabanditka [31]3 years ago

7 0

Six.

If you want to double the force, you need to double either v or B. Changing the angle does not give totally predictable results. The formula is

F = q*v*B Sin(theta) altering theta isn't linear. Doubling B is. The answer is A

Seven

Seven is correct. The poles on a magnet is where the field is the strongest.

Eight

I can't really answer this question. Much depends on where you are and whether or not the readings are true or erratic. It is taking only one parameter into account where in reality there are many. My guess would be that as you move up the longitude the angle increases. So the answer would be near but not at the north pole. Seek out your notes for confirmation.

You might be interested in

The element chlorine has two naturally occurring isotopes. About 75% of chlorine isotopes are Cl-35 and about 25% are Cl-37.

Mila [183]

ur answer i would beileve is a

8 0

3 years ago

Read 2 more answers

Plsss help I don’t understand this

nekit [7.7K]

B is the correct option.

1. Given eqn;

S(t) = 1/2t² - 4t + 8

2.Differentiate the above eqn with respect to t;

<u>d(S(t))</u> = t - 4

When distance, S, is differentiated it results to velocity.

V = t - 4

at t = 10

V = 10 - 4

V = 6 feet/s

6 0

2 years ago

Read 2 more answers

A purse at radius 2.30 m and a wallet at radius 3.45 m travel in uniform circular motion on the floor of a merry-go-round as the

ivolga24 [154]

Answer:

The acceleration of the wallet is $3\hat{i}+6\hat{j}$

Explanation:

Given that,

Radius of purse r= 2.30 m

Radius of wallet r'= 3.45 m

Acceleration of the purse $a=2\hat{i}+4.00\hat{j}$

We need to calculate the acceleration of the wallet

Using formula of acceleration

$a=r\omega^2$

Both the purse and wallet have same angular velocity

$\omega=\omega'$

$\sqrt{\dfrac{a}{r}}=\sqrt{\dfrac{a'}{r'}}$

$\dfrac{a}{r}=\dfrac{a'}{r'}$

$\dfrac{a'}{a}=\dfrac{r'}{r}$

$\dfrac{a'}{a}=\dfrac{3.45}{2.30}$

$\dfrac{a'}{a}=\dfrac{3}{2}$

$a'=\dfrac{3}{2}\times(2\hat{i}+4.00\hat{j})$

$a'=3\hat{i}+6\hat{j}$

Hence, The acceleration of the wallet is $3\hat{i}+6\hat{j}$

4 0

3 years ago

A bird flies overhead from where you stand at an altitude of 270.0ĵ m and at a velocity horizontal to the ground of 14.0î m/is.

Varvara68 [4.7K]

Answer:

L = 8694 Kg.m²/s

Explanation:

r = 270 ĵ m

v = 14 î m/s

m = 2.3 kg

θ = 90º

L = ?

We can apply the equation

L = m*v*r*Sin θ

L = (2.3 kg)*(14 m/s)*(270 m)*Sin 90º = 8694 Kg.m²/s

8 0

3 years ago

An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl

svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Displacement, s = 192 m

Acceleration, a = 9.81 m/s²

Substituting

s = ut + 0.5 at²

192 = 0 x t + 0.5 x 9.81 x t²

t = 6.26 seconds

Now we need to find horizontal distance traveled in 6.26 seconds by the package.

We have equation of motion, s = ut + 0.5 at²

Initial velocity, u = 41 m/s

Time, t = 6.26 s

Acceleration, a = 0 m/s²

Substituting

s = ut + 0.5 at²

s = 41 x 6.26 + 0.5 x 0 x 6.26²

s = 256.52 m

The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

Initial velocity, u = 0 m/s

Time, t = 6.26 s

Acceleration, a = 9.81 m/s²

Substituting

v = u+ at

v = 0 + 9.81 x 6.26 = 61.41 m/s

Vertical component of velocity = 61.41 m/s

4 0

3 years ago