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2 years ago

Can y’all help me with 5 plsssss

Physics

2 answers:

Lemur [1.5K]2 years ago

6 0

Answer:

Bar graph would work :)

Alona [7]2 years ago

4 0

Answer:

Bar graph

Explanation:

each day collects data so a bar graph would work.

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An experiment is conducted on a long straight wire of diameter d. A constant current is sent through the wire and the magnetic f

Anton [14]

Answer:

$B_2 = \frac{B_1}{2}$

Explanation:

As per Ampere's law the magnetic field at the surface of the wire is given as

$\int B. dL = \mu_0 i$

here we have

$B . (\pi d) = \mu_0 i$

so we will have

$B_1 = \frac{\mu_0 i}{\pi d}$

now again we use same value of current but wire with double the diameter

so the magnetic field at the surface is given as

$B_2 = \frac{\mu_0 i}{2\pi d}$

so we have

$B_2 = \frac{B_1}{2}$

5 0

3 years ago

a person exerts a force of 55 n on the end of a door 74 cm wide. what is the magnitude of the torque if the force is exerted?

ratelena [41]

Answer: Torque T = 40.7 Nm

Explanation:

Torque or turning effect is the rotational equivalent of linear force, i.e moment of force about a point.

Given that the force is applied perpendicular to the edge (angle 90°)

Torque can be represented by T;

T = FrSin∅ = Fr ( sin90 = 1)

T = force × distance

T = F × r

F = 55N

r = 74cm = 0.74m

T = 55N × 0.74m

T = 40.7 Nm

5 0

3 years ago

A car (m = 1302 kg) traveling along a road begins accelerating with a constant acceleration of 1.50 m/s2 in the direction of mot

antiseptic1488 [7]

First we will use the concepts of motion kinetics for which the final speed is defined as shown below,

$v_f^2=v_i^2+2as$

Here,

$v_f$ = Final velocity

$v_i$ = Initial velocity

a = Acceleration

s = Distance

Replacing,

$(35)^2 = v_i^2+2(1.5)(392)$

$v_i = 7m/s$

Using the conservation of energy for kinetic energy we have,

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

$KE = \frac{1}{2}(1302)(7)^2$

$KE = 31900J$

Therefore the kinetic energy of the car is 31900J

5 0

3 years ago

A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:

olya-2409 [2.1K]

To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.

PART A ) The Impulse can be calculcated as follows

$L= F\Delta t$

Where,

F = Force

$\Delta t =$ Change in time

Replacing,

$L = (35N)(0.2s)$

$L= 7N\cdot s$

PART B) At the same time the momentum follows the conservation of momentum where:

Initial momentum= Final momentum

And the change in momentum is equal to the Impulse, then

$\Delta p = L$

And

$\Delta p = p_f - p_i$

There is not initial momentum then

$\Delta p = p_f$

$L = p_f$

$p_f = 7N\cdot s = 7kg\cdot m/s$

8 0

3 years ago

A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>

6 0

3 years ago