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kenny6666 [7]
3 years ago
12

The current loop lies flat on a horizontal surface. Facing the loop from the top, the current moves in a clockwise direction. Wh

ich way does the magnetic field point outside the loop?
a) The field points down.
b) The field points down.
c) The field points to the left.
d) The field points up
e) The field points to the right
Physics
1 answer:
lapo4ka [179]3 years ago
6 0

Answer:

e

Explanation:

the field points to the right

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1. You wish to heat 20 kg of water from 40°C to 80°C. How many kcal of heat are necessary to do this? To how many kJ does this c
adelina 88 [10]

Answer:

<h2>3,343.68kJ </h2>

Explanation:

Heat energy used up can be calculated using the formula:

H = mcΔt

m = mass oof the object (in kg) = 20kg

c = specific heat capacity of water = 4179.6J/kg°C

Δt change in temperature = 80-40 = 40°C

H= 20* 4179.6 * 40

H = 3,343,680Joules

H = 3,343.68kJ

8 0
3 years ago
If m represent mass in kg, v represents speed in m/s and r represents radius in m show F in the formula F= (mv^2)/r can be expre
Dmitrij [34]
M <span>represent mass in kg
</span><span>v represents speed in m/s
</span><span>r represents radius in m

Now, just substitute these into the formula:
</span>F =  \frac{m* v^{2} }{r} =\frac{kg* ( \frac{m}{s} )^{2} }{m} =\frac{kg* \frac{m^{2}}{s^{2}} }{m} = \frac{kg*m^{2}}{s^{2}*m } =\frac{kg*m}{s^{2} }<span>

</span>
3 0
3 years ago
A 20 kg truck drives in a circle of radius 4 m at 10m/s. What is the centripetal acceleration of the truck?
asambeis [7]

Answer:

B. 25 m/s/s

Explanation:

Centripetal acceleration is the square of the tangential velocity divided by the radius of curvature.

a = v² / r

Given v = 10 m/s and r = 4 m:

a = (10 m/s)² / 4 m

a = 25 m/s²

4 0
3 years ago
Question is attatched!
lina2011 [118]
All it does is lets him pull in a more convenient direction to raise the load. It has no effect on the required force.
4 0
3 years ago
A wheel that is rotating at 33.3 rad/s is given an angular acceleration of 2.15 rad/s 2. Through what angle has the wheel turned
Neporo4naja [7]

Answer:

The angle through which the wheel turned is 947.7 rad.

Explanation:

initial angular velocity, \omega _i = 33.3 rad/s

angular acceleration, α = 2.15 rad/s²

final angular velocity, \omega_f = 72 rad/s

angle the wheel turned, θ = ?

The angle through which the wheel turned can be calculated by applying the following kinematic equation;

\omega_f^2 = \omega_i^2 + 2\alpha \theta\\\\\theta = \frac{\omega_f^2\   -\  \omega_i^2}{2\alpha } \\\\\theta = \frac{(72)^2\   -\  (33.3)^2}{2(2.15)}\\\\\theta = 947.7 \ rad

Therefore, the angle through which the wheel turned is 947.7 rad.

5 0
3 years ago
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