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Karo-lina-s [1.5K]

3 years ago

15

Newtons second law states that Force=Mass × Acceleration.The acceleration of gravity in 9.8 m/s.How much force is gravity exerti

ng on the 100g ball

1 answer:

Naddika [18.5K]3 years ago

5 0

.98 Newton’s because you convert 100 g to kg which is .1 kg them you multiply.1 kg by 9.8 and get .98 and the units of the force are in Newton’s

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The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

or

Time = $\frac{\textup{0.016}}{\textup{23}}$

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

or

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

or

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s

trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum $p_{i}$ must be equal to the final momentum $p_{f}$ , and taking into account this is aninelastic collision:

Before the collision:

$p_{i}=mV_{o}+MU_{o}$ (1)

After the collision:

$p_{f}=(m+M)V_{f}$ (2)

Where:

$m=1380 kg$ is the mass of the car

$V_{o}=23 m/s$ is the velocity of the car, directed to the north

$M=1625 kg$ is the mass of the truck

$U_{o}=-26 m/s$ is the velocity of the truck, directed to the south

$V_{f}$ is the final velocity of both the car and the truck

$p_{i}=p_{f}$ (3)

$mV_{o}+MU_{o}=(m+M)V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{mV_{o}+MU_{o}}{m+M}$ (5)

$V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg}$ (6)

Finally:

$V_{f}=-3.49 m/s$ The negative sign indicates the direction of the velocity is to the south

8 0

3 years ago

Toy car in a science experiment covers 1.6 meters in half a second. If a the car travels at a steady speed, how far will it go i

Tanzania [10]

The answer is D. 32 m.

The simple equation that connects speed (v), time (t), and distance (d) can be expressed as:
$v= \frac{d}{t}$ ⇒ $d=v*t$

It is given:
$v = \frac{1.6m}{0.5s} = \frac{1.6m*2}{0.5s*2}= \frac{3.2m}{1s} = 3.2 m/s$
t = 10 s
d = ?

So:
$d= v*t=3.2m/s*10s = 32m$

3 0

3 years ago

Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.

solong [7]

Answer:

The answer is D.

7 0

3 years ago

Read 2 more answers

A physics professor is pushed up a ramp inclined upward at 30.0° above the horizontal as she sits in her desk chair, which slide

11111nata11111 [884]

Answer:

V = 3.17 m/s

Explanation:

Given

Mass of the professor m = 85.0 kg

Angle of the ramp θ = 30.0°

Length travelled L = 2.50 m

Force applied F = 600 N

Initial Speed u = 2.00 m/s

Solution

Work = Change in kinetic energy

$F_{net}d = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}\\\frac{2F_{net}d }{m} = v^{2} -u^{2}\\ v^{2} =\frac{2F_{net}d }{m} +u^{2}\\ v^{2} =\frac{2(600cos30 - 85\times 9.8 \times sin30) \times 2.5 }{85} +2.00^{2}\\ v^{2} = 10.066\\v = 3..17m/s$

7 0

3 years ago