What is the spring constant for a supermarket scale that stretches 0.01 m when a force of 4 N is applied

Pls answer this<br> quick as possible

mihalych1998 [28]

Answer:

The answer is the 4th one

Explanation:

Good luck man

7 0

3 years ago

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.

snow_lady [41]

Complete Question

A uniform electric field of magnitude 144 kV/m is directed upward in a region of space. A uniform magnetic field of magnitude 0.38 T perpendicular to the electric field also exists in this region. A beam of positively charged particles travels into the region. Determine the speed of the particles at which they will not be deflected by the crossed electric and magnetic fields. (Assume the beam of particles travels perpendicularly to both fields.)

Answer:

The velocity is $v = 3.79 *10^{5} \ m/s$

Explanation:

From the question we are told that

The magnitude of the electric field is $E = 144 \ kV /m = 144*10^{3} \ V/m$

The magnetic field is $B = 0.38 \ T$

The force due to the electric field is mathematically represented as

$F_e = E * q$

and

The force due to the magnetic field is mathematically represented as

$F_b = q * v * B * sin(\theta )$

Now given that it is perpendicular , $\theta = 90$

=> $F_b = q * v * B * sin(90)$

=> $F_b = q * v * B$

Now given that it is not deflected it means that

$F_ e = F_b$

=> $q * E = q * v * B$

=> $v = \frac{E}{B }$

substituting values

$v = \frac{ 144 *10^{3}}{0.38 }$

$v = 3.79 *10^{5} \ m/s$

7 0

3 years ago

Amanda goes shopping for light bulbs. In the light bulb aisle, she finds the following incandescent light bulbs. Can you help he

Luden [163]

Answer:

The least efficient light bulb is the first one (25 W - 210 Lumen)

Explanation:

Efficiency can be defined as what you want to obtain over what you need to produce it. In this case Eff= Wattage / Lumen. For each light bulb, their efficiency is: 8.4 / 11.12 / 15.7 Lum/W

4 0

3 years ago

A commercial jet liner takes off with an average acceleration of 3 g. How long does it take to reach the end of its runway which

ANEK [815]

Answer:

The time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

Explanation:

Given;

average acceleration of the commercial Jet liner, a = 3g = 3 x 9.8 m/s² = 29.4 m/s²

distance traveled by the commercial Jet liner, s = 1542 m

The time taken for the commercial Jet liner to reach the end of its runway is calculated as follows;

s = ut + ¹/₂at²

where;

u is the initial velocity of the commercial Jet liner = 0

s = 0 + ¹/₂at²

s = ¹/₂at²

2s = at²

$t = \sqrt{\frac{2s}{a} } \\\\t = \sqrt{\frac{2 \times 1524}{29.4} } \\\\t = 10.18 \ s$

Therefore, the time taken for the commercial Jet liner to reach the end of its runway is 10.18 s.

6 0

3 years ago

A simple pendulum consists of a small object of mass 1.70 kg hanging at the end of a 2.50 m long light string that is connected

kompoz [17]

Answer: 3.63 Nm

Explanation:

from the question we were given the following

mass = 1.7 kg

length of the string (r) = 2.5 ,m

angle θ = 5 degrees

acceleration due to gravity (g) = 9.8 m/s

we can calculate the torque using the formula Torques = m x g x r x sin θ

torque = 1.7 x 9.8 x 2.5 x sin 5

torque = 3.63 Nm

8 0

3 years ago