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3 years ago

PLEASE HELP:::

Physics

2 answers:

vladimir1956 [14]3 years ago

6 0

Newton's first law of motion is often stated as

<span>An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force. Therefore, the answer here is A) Newton's First Law.

Hope this helped ;)
</span>

OleMash [197]3 years ago

6 0

A). Newtons first Law

You might be interested in

Hi there hope your having a great day!! my questions both are SCIENCE laws of motion related fyi

Ann [662]

1. 6.0 m

force/mass× acceleration is the equation

since the accelaration is unknown you would divide 180 by the mass of 30kg which is 6

2.A

4 0

3 years ago

A bowler throws a bowling ball of radius R = 11 cm along a lane. The ball slides on the lane with initial speed <img src="https:

s344n2d4d5 [400]

Answer:

a) v_com= Rω

b) -2.254 m/ $s^{2}$

c) 51.2 rad/ $s^{2}$

d) t=1.08 seconds

e) x=7.865m

f) v_roll=6.07m

Explanation:

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse

$\alpha$ =(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2

=7.865m

v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

=6.07m/sec

Initially, the ball is travelling with v_com=v_0

Wen not rotating, at the initial stage the ball must be sliding along the surface.

This motion therefore generates a frictional force F_r at the point of contact.

a) Let the velocity at the point of contact be v_bottom

v_bottom=v_com-Rω

Therefore when ω=0, v_bottom=v_com

So when the ball begins rolling

v_com= Rω

b) F_r=μ_rmg

〖-F〗_r=ma_com

a_com=(〖-μ〗_r mg)/m

a_com=-μ_rg

a_com=-(0.23)(9.8)

a_com=-2.254m/s^2

Te negative sow decrearse

c) α=(μ_r mgR)/I = (〖5μ〗_r mgR)/2mRR

=(〖5μ〗_r g)/2R

=(5*(0.23)*(9.8))/(2*0.11)

=51.2 rad/s^2

d) t=v_0/(〖-a〗_com+Rα)

=8.5/(2.255+0.11*(51.2))

=8.5/7.886

=1.08 seconds

e) X=v_0 t+1/2 a_com t^2

X=8.5*(2.254) - 1/2 (2.254)*〖1.08〗^2

=7.865m

f) v_roll=v_0+a_com t_r

=8.5-(2.254)(1.08)

=6.07m/sec

7 0

3 years ago

_______________have a negative charge and are located on the outside of the nucleus.

GuDViN [60]

May 5th is Adachi day change your profile picture

6 0

3 years ago

For the wave of light you generated in the Part B, calculate the amount of energy in 1.0 mol of photons with that same frequency

Angelina_Jolie [31]

Answer:

2.7 J

Explanation:

The energy of one photon is given by

$E=hf$

where

h is the Planck constant

f is the frequency

For the photons in this problem,

$f=6.8\cdot 10^9 Hz$

So the energy of one photon is

$E_1=(6.63\cdot 10^{-34})(6.8\cdot 10^9 )=4.5\cdot 10^{-24} J$

The number of photons contained in 1.0 mol is

$N_A = 6.022\cdot 10^{23} mol^{-1}$ (Avogadro number)

So the total energy of $N_A$ photons contained in 1.0 mol is

$E=N_A E_1 =(6.022\cdot 10^{23})(4.5\cdot 10^{-24})=2.7 J$

3 0

3 years ago

The diver is on a board3.00m above the water. She jumps straight up at 3.25m/s. How many seconds later does she hit the water.

tigry1 [53]

Speed =distance/time
3.25=3.00/time
3.25xt=3.00
t=3/3.25
s=0.9s

5 0

3 years ago