Ask question

Ask question

All categories

3 years ago

15

Luis and Aisha conducted an experiment. They exerted different forces on four objects. Their results are shown in the table.

2 answers:

lesya692 [45]3 years ago

4 0

Answer:

Object 3 has greatest acceleration.

Explanation:

Objects Mass Force

1 10 kg 4 N

2 100 grams 20 N

3 10 grams 4 N

4 1 kg 20 N

Acceleration of object 1,

$a_1=\dfrac{F_1}{m_1}\\\\a_1=\dfrac{4}{10}\\\\a_1=0.4\ m/s^2$

Acceleration of object 2,

$a_2=\dfrac{F_2}{m_2}\\\\a_2=\dfrac{20}{0.1}\\\\a_2=200\ m/s^2$

Acceleration of object 3,

$a_3=\dfrac{F_3}{m_3}\\\\a_3=\dfrac{4}{0.01}\\\\a_3=400\ m/s^2$

Acceleration of object 4,

$a_4=\dfrac{F_4}{m_4}\\\\a_4=\dfrac{20}{1}\\\\a_3=20\ m/s^2$

It is clear that the acceleration of object 3 is $400\ m/s^2$ and it is greatest of all. So, the correct option is (3).

Tasya [4]3 years ago

3 0

Answer:

Object 3

Explanation:

You might be interested in

Grasses, shrubs and trees are called producers because they make ______a. water_____b. carbon dioxide_____c. minerals_____d. foo

LekaFEV [45]

B?
Carbon Dioxid? Or is is c?

4 0

3 years ago

Read 2 more answers

How much power does it take to lift 70.0 N to 5.0 m high in 5.00 s?

lesantik [10]

Answer:

Power = 70 W

Explanation:

Given that,

Force, F = 70 N

Height, h = 5 m

Time, t = 5 s

We need to find the power of the object. We know that,

Power = work done/time

Put all the values,

$P=\dfrac{Fd}{t}\\\\P=\dfrac{70\times 5}{5}\\\\P=70\ W$

So, the required power is 70 W.

3 0

2 years ago

Pablo's engineering team has developed a new material that is strong enough to withstand major impacts, including bullets, witho

marin [14]

Answer:

the answer is c

Explanation:

4 0

3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

6 0

3 years ago

WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di

Yuki888 [10]

Answer:

a

$\lambda = 202.7 \ m$

b

$w = 9.3 *10^{6} \ rad/s$

c

$k = 0.031 m^{-1}$

d

$E_{max} = 9.0 *10^{-3} \ V/m$

Explanation:

From the question we are told that

The frequency of the radio station is $f= 1480 \ kHz = 1480 *10^{3}\ Hz$

The magnitude of the magnetic field is $B = 3.0* 10^{-11} \ T$

Generally the wavelength is mathematically represented as

$\lambda = \frac{c}{f}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$\lambda = \frac{3.0 *10^{8}}{ 1480 *10^{3}}$

=> $\lambda = 202.7 \ m$

Generally the angular frequency is mathematically represented as

$w = 2 \pi * f$

=> $w = 2 * 3.142 * 1480 *10^{3}$

=> $w = 9.3 *10^{6} \ rad/s$

Generally the wave number is mathematically represented as

=> $k = \frac{2 \pi }{\lambda}$

=> $k = \frac{2 * 3.142 }{ 202.7}$

=> $k = 0.031 m^{-1}$

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

$E_{max} = c * B$

=> $E_{max} = 3.0 *10^{8} * 3.0* 10^{-11}$

=> $E_{max} = 9.0 *10^{-3} \ V/m$

3 0

2 years ago