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3 years ago

TRUE OR FALSE, I would be so happy pzl help0 me

Physics

2 answers:

Darina [25.2K]3 years ago

6 0

sorry but i only know of 7 and 8 and the both are true

Art [367]3 years ago

5 0

Answer:

hhh...Are you in 2nd grade? Im pretty sure my lil sister knows this

Explanation:

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A car drives over a hilltop that has a radius of curvature 0.120 km at the top of the hill. At what speed would the car be trave

Mashutka [201]

Answer:

34.3 m/s

Explanation:

Newton's Second Law states that the resultant of the forces acting on the car is equal to the product between the mass of the car, m, and the centripetal acceleration $a_c$ (because the car is moving of circular motion). So at the top of the hill the equation of the forces is:

$mg-R = m a_c = m\frac{v^2}{r}$

where

(mg) is the weight of the car (downward), with m being the car's mass and g=9.8 m/s^2 is the acceleration due to gravity

R is the normal reaction exerted by the road on the car (upward, so with negative sign)

v is the speed of the car

r = 0.120 km = 120 m is the radius of the curve

The problem is asking for the speed that the car would have when it tires just barely lose contact with the road: this means requiring that the normal reaction is zero, R=0. Substituting into the equation and solving for v, we find:

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(120 m)}=34.3 m/s$

6 0

3 years ago

Henri Becquerel is credited with the discovery of:

Yuri [45]

Radium........ Hope this helps :)

5 0

3 years ago

Read 2 more answers

The substances in the table are combined, and Substance 1 loses 40 calories of heat. How many calories of heat will Substance 2

Luda [366]

Over time, the substances will reach equilibrium, meaning the heat calories lost by Substance 1 will be gained by Substance 2. Therefore, Substance 2 will eventually gain 40 calories of heat.

5 0

3 years ago

Read 2 more answers

A point charge gives rise to an electric field with magnitude 2 N/C at a distance of 4 m. If the distance is increased to 20 m,

worty [1.4K]

Answer:

0.08 N/C

Explanation:

Electric Field: This is defined as the force per unit charge exerted at a point. The expression for electric field is given as,

E = Kq/r².............................. Equation 1

Where E = Electric Field, q = Charge, k = proportionality constant, r = distance.

making q the subject of the equation,

q = Er²/k............................... Equation 2

Given: E = 2 N/C, r = 4 m,

Substitute into equation 2

q = 2(4)²/k

q = 32/k C.

When r is increased to 20 m,

E = k(32/k)/20²

E = 32/400

E = 0.08 N/C.

Hence the electric Field = 0.08 N/C

5 0

4 years ago

A 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if t

Arisa [49]

A) $2.4\cdot 10^{-16}kg$

The radius of the oil droplet is half of its diameter:

$r=\frac{d}{2}=\frac{0.80 \mu m}{2}=0.40 \mu m = 0.4\cdot 10^{-6}m$

Assuming the droplet is spherical, its volume is given by

$V=\frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.4\cdot 10^{-6} m)^3=2.68\cdot 10^{-19} m^3$

The density of the droplet is

$\rho=885 kg/m^3$

Therefore, the mass of the droplet is equal to the product between volume and density:

$m=\rho V=(885 kg/m^3)(2.68\cdot 10^{-19} m^3)=2.4\cdot 10^{-16}kg$

B) $1.5\cdot 10^{-18}C$

The potential difference across the electrodes is

$V=17.8 V$

and the distance between the plates is

$d=11 mm=0.011 m$

So the electric field between the electrodes is

$E=\frac{V}{d}=\frac{17.8 V}{0.011 m}=1618.2 V/m$

The droplet hangs motionless between the electrodes if the electric force on it is equal to the weight of the droplet:

$qE=mg$

So, from this equation, we can find the charge of the droplet:

$q=\frac{mg}{E}=\frac{(2.4\cdot 10^{-16}kg)(9.81 m/s^2)}{1618.2 V/m}=1.5\cdot 10^{-18}C$

C) Surplus of 9 electrons

The droplet is hanging near the upper electrode, which is positive: since unlike charges attract each other, the droplet must be negatively charged. So the real charge on the droplet is

$q=-1.5\cdot 10^{-18}C$

we can think this charge has made of N excess electrons, so the net charge is given by

$q=Ne$

where

$e=-1.6\cdot 10^{-19}C$ is the charge of each electron

Re-arranging the equation for N, we find:

$N=\frac{q}{e}=\frac{-1.5\cdot 10^{-18}C}{-1.6\cdot 10^{-19}C}=9.4 \sim 9$

so, a surplus of 9 electrons.

3 0

4 years ago