Ask question

Ask question

All categories

3 years ago

15

A stone of mass m is thrown upwards at an angle φ. The moment the stone leaves your hand it has an acceleration (friction is neg

ligible) equal to
g

g sin (φ)

zero

1 answer:

Lelechka [254]3 years ago

5 0

The correct answer is:

= g

Explanation:

When the object leaves the hand it has a force acted on it mainly it’s weight.

You might be interested in

56 kg girl slides down the frictionless slip-n-slide at a constant speed of 35 m/s. What is the kinetic energy in Joules of the

melisa1 [442]

Answer:

93 joules

Explanation:

7 0

3 years ago

A potential difference of 1.20 V will be applied to a 33.0 m length of 18-gauge copper wire (diameter = 0.0400 in.). Calculate (

Nostrana [21]

Answer:A) Current = 1.739A, B)current density, J = 2.147x10^6 A/m2

magnitude of electric field , E = 0.036 N/C

)rate of thermal energy, P =2.086W

Explanation:

Resistance = R = ρL/A

But the cross-section area of the wire. is given as

Diameter / 2 = 0.04/2 =0.02in to m = 0.02 / 39.37= 0.000508

A = πr^2 = π x 0.000508^2 = 8.10 x 10^-7

since resistivity of copper,ρ= 17x10-9 ohm.m

so resistance is R = ρL/A

17x10-9 x 33 / 8.1x10-7

= 0.69 ohm.

A) Current = I = Voltage /Resistance =1.20/0.69 =1.739A

B)current density, J = Current /Area

= 1.739/8.1x10-7

= 2.147x10^6 A/m2

c)magnitude of electric field , E = Current density x resistivity =J ρ

E = 2.147 x 10^6 x 17 x 10^-9

E = 0.036 N/C

D)rate of thermal energy, P = I² R =1.739² X 0.69

=2.086W

6 0

3 years ago

A deuteron particle consists of one proton and one neutron and has a mass of 3.34x10^-27 kg. A deuteron particle moving horizont

nikklg [1K]

_dThe radius of curvature of a subatomic particle under a magnetic field is given by the following formula:

$r=\frac{mv}{qB}$

Where:

$\begin{gathered} r=\text{ radius} \\ v=\text{ velocity} \\ q=\text{ charge} \\ B=\text{ magnetic field} \end{gathered}$

We can determine the quotient between the velocity and the charge of the deuteron particle from the formula. First, we divide both sides by the mass:

$\frac{r_d}{m_d}=\frac{v}{q_B_}$

Now, we multiply both sides by the magnetic field "B":

$\frac{Br_d}{m_d}=\frac{v}{q}$

Since the charge of the deuterion is the same as the charge of the proton and the velocity we are considering are the same this means that the quotient between velocity and charge is the same for both particles. Therefore, we can apply the formula for the radius again, this time for the proton:

$r_p=\frac{m_pv}{qB}$

And substitute the quotient between velocity and charge:

$r_p=\frac{m_p}{B}(\frac{Br_d}{m_d})$

Now, we cancel out the magnetic field:

$r_p=\frac{m_pr_d}{m_d}$

Now, we substitute the values:

$r_p=\frac{(1.67\times10^{-27}kg)(0.385m)}{(3.34\times10^{-27}kg)}$

Solving the operations:

$r_p=0.193m=19.3cm$

Therefore, the radius is 19.3 cm.

3 0

1 year ago

A student stands on the edge of a cliff that is 300 m high and kicks a rock horizontally. 7.8 seconds later, the rock hits the g

mart [117]

Answer:

6.65m/s

Explanation:

Using the equation of motion

S = ut + 1/2gt²

S is the height of fall

t is the time

u is the horizontal velocity

g is the acceleration due to gravity

Given

S = 300 + 50

S = 350m

t = 7.8seconds

g = 9.8m/s^2

Get S

S = 7.8u + 1/2(9.8)(7.8)²

S = 7.8u + 298.116

350 = 7.8u + 298.116

7.8u = 350 - 298.116

7.8u = 51.884

u = 51.884/7.8

u = 6.65m/s

Hence the rock's horizontal velocity was 6.65m/s

6 0

3 years ago

A student leaves a pencil on a desk and the pencil stays in the same spot until another student picks it up. Which law would tha

atroni [7]

The first law

In Newtons first law it states that an object will not change its motion unless a force acts upon it.

5 0

3 years ago