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marissa [1.9K]
3 years ago
14

I need help with science please, ill give brianlist’s :

Physics
2 answers:
Elis [28]3 years ago
7 0

Answer:

true , true, and I'm not sure about the last one!

Explanation:

evablogger [386]3 years ago
5 0
The answer Is True!!! Good luck
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You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn f
Verdich [7]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. We will calculate the initial velocity of the object, and from it, we will calculate the final position. With the considerations made in the statement we will obtain the total height. Initial velocity of the acorn,

u = 0m/s

Also, it is given that the acorn takes 0.201s to pass the length of the meter stick.

s = ut+\frac{1}{2} at^2

Replacing,

1 = u(0.141)+ \frac{1}{2} (9.8)(0.141)^2

u =6.4013m/s

The height of the acorn above the meter stick can be calculated as,

v^2 = u^2 +2gh

h = \frac{v^2-u^2}{2g}

h = \frac{6.4013^2-0^2}{2(9.8)}

h = 2.0906m

Also the top of the meter stick is 1.87m above the ground hence the height of the acorn above the ground is

h = 2.0906+1.87

h = 3.9606m

4 0
4 years ago
A 6.0 kg box slides down an inclined plane that makes an angle of 39° with the horizontal. If the coefficient of kinetic frictio
Blababa [14]

Answer:

(B) 1.6 m/s^2

Explanation:

The equation of the forces acting on the box in the direction parallel to the slope is:

mg sin \theta - \mu N = ma (1)

where

mg sin \theta is the component of the weight parallel to the slope, with m = 6.0 kg being the mass of the box, g = 9.8 m/s^2 being the acceleration of gravity, \theta=39^{\circ} being the angle of the incline

\mu N is the frictional force, with \mu = 0.6 being the coefficient of kinetic friction, N being the normal reaction of the plane

a is the acceleration

The equation of the force along the direction perpendicular to the slope is

N-mg cos \theta =0

where mg cos \theta is the component of the weight in the direction perpendicular to the slope. Solving for N,

N=mg cos \theta

Substituting into (1), solving for a, we find the acceleration:

a=gsin \theta- \mu g cos \theta=(9.8)(sin 39^{\circ})-(0.6)(9.8)(cos 39^{\circ})=1.6 m/s^2

6 0
4 years ago
Okay so I've have 10 Brainliest, through my whole time using brainly- I want people to get brainly and points so Im giving away
Nookie1986 [14]

answer:

thank youu for the points cute drawing by the way (✿◠‿◠)

when did you join brainly though?

have a wonderful dayyy !!

4 0
3 years ago
Read 2 more answers
Two 10-cm-diameter charged rings face each other, 23.0 cm apart. Both rings are charged to 40.0 nC . What is the electric field
Volgvan

Answer: Two 10-cm-diameter charged rings face each other, 25.0cm apart. Both rings are charged to +20.0nC.

Explanation:

4 0
3 years ago
A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be co
Naddika [18.5K]

Answer:

a = 45 m/s/s

Explanation:

As we know that total mass of the rocket is

M = 6000 kg

total mass of the fuel is given as

m = 2000 kg

all the fuel is burnt in 15 s

so rate of the fuel burning is given as

\frac{dm}{dt} = \frac{2000}{15}

\frac{dm}{dt} = 133.33 kg/s

now the thrust force on the rocket is given as

F_{th} = v\frac{dm}{dt}

(6000 - 133.33 t) a = (900 + v)(133.33)

so we have

\frac{dv}{900 + v} = (133.33)\frac{dt}{6000 - 133.33 t}

so we have

ln(\frac{900 + v}{900}) = - ln(\frac{6000 - 133.3 t}{6000})

1 + \frac{v}{900} = \frac{6000}{6000 - 133.3 t}

now acceleration is rate of change in velocity

\frac{1}{900}\frac{dv}{dt} = \frac{133.3\times 6000}{(6000 - 133.3t)^2}

so acceleration at t = 15 s

a = 45 m/s^2

3 0
4 years ago
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