Which two features are part of a line graph

A 3.0kg mass tied to a string

dem82 [27]

Answer:

$\boxed{\sf Tension \ in \ the \ string \ (T) = 3 \ kN}$

Given:

Mass (m) = 3.0 kg

Uniform speed (v) = 20 m/s

Length of string (r) = 40 cm = 0.4 m

To Find:

Tension in the string (T)

Explanation:

Tension (T) is the string will be equal to centripetal force ( $\sf F_c$ ).

$\boxed{ \bold{ T = F_c = \frac{m {v}^{2} }{r} }}$

Substituting value of m, v & r in the equation:

$\sf \implies T = \frac{3 \times {20}^{2} }{0.4} \\ \\ \sf \implies T = \frac{3 \times 400}{0.4} \\ \\ \sf \implies T =3 \times 1000 \\ \\ \sf \implies T =3000 \: N \\ \\ \sf \implies T =3 \: kN$

$\therefore$

Tension in the string (T) = 3 kN

5 0

3 years ago

A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.

cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

$a=1.5 m/s^2$

So we can re-arrange the equation above to find their combined mass:

$m=\frac{F}{a}=\frac{75}{1.5}=50 kg$

3 0

3 years ago

7. How is the current found in a parallel circuit?

Naya [18.7K]

Answer:

the current in a parallel circuit is found by idk

Explanation:

4 0

3 years ago

Under which type of change would more organisms be able to survive? Why?

topjm [15]

The adaptation that is genetic change or mutatiton help makes an organism to survive better in their new environment.

8 0

3 years ago

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle

ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E = 5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0

3 years ago