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3 years ago

Sam shared with his class that he lives on Oak Street. Sam is likely how old?

Physics

2 answers:

Lyrx [107]3 years ago

4 0

D FOUR years old hope this helps

mark brainliest please

yulyashka [42]3 years ago

4 0

A child will be able to understand and recognize this by the age of four years old.

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Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder

kramer

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

$U = \frac{1}{2}kx^2$

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

$KE = \frac{1}{2}kx^2$

here we have

k = 70 N/m

x = 0.4 m

$KE = \frac{1}{2}(70)(0.4)^2$

$KE = 5.6 J$

Part c)

Now to find the speed we know that

$KE = \frac{1}{2} mv^2$

$5.6 = \frac{1}{2}0.3v^2$

$v = 6.11 m/s$

so its speed is 6.11 m/s

8 0

3 years ago

Read 2 more answers

A box is being pulled by two ropes. Eduardo pulls to the left with a force of 500 N, and Clara pulls to the right with a force o

ZanzabumX [31]

The box is moving in the direction of Eduardo with a force of 300N since he is pulling 300N stronger than Clara.

8 0

3 years ago

Read 2 more answers

A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)

egoroff_w [7]

Explanation:

The given data is as follows.

q = 6.0 nC = $6 \times 10^{-9} C$

inner radius (r) = 1.0 cm = 0.01 m (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

$\sigma = \frac{q_{in}}{A}$ .......... (1)

Since, area of the sphere is as follows.

A = $4 \pi r^{2}$ ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

$\sigma = \frac{q_{in}}{4 \pi r^{2}}$

= $\frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}$

= $0.477 \times 10^{-5}$

or, = 4.77 $\mu C/m^{2}$

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 $\mu C/m^{2}$ .

5 0

3 years ago

In 1993, Cuban athlete Javier Sotomayor set the world record for the high jump. The gravitational potential energy associated wi

Katarina [22]

Answer:2.541

Explanation:

Well , Potential Energy = mgh

m=mass = 82

g=acceleration of gravity=9.80m/s^2

h=what we are looking for

PE=mgh

PE/(mg) = h

Substitute in the values:

1970/(82 x 9.8) = h 2.541

5 0

3 years ago

particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago