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Serhud [2]
3 years ago
14

Two men decide to use their cars to pull a

Physics
1 answer:
Ulleksa [173]3 years ago
7 0
The resultant force in the direction the truck is headed is:

734N*cos(31) + 1084N*cos(23)
629N+998N = 1627N
You might be interested in
The particles that make up protons and neutrons are called
Sphinxa [80]

Answer:

Quark

Explanation:

In the past the proton,neutron and electron were believed to be indivisible. Quarks were later discovered to be particles that make up the proton and neutron. Quarks are known as fast moving points of energy and are usually joined to one another through gluons. Neutron has two down quarks and one up quark while proton has two up quarks and one down quark.

6 0
3 years ago
Two Force one of 12 Newton and another 24 Newton acts at 90 degree with each other find the resultant of two force and its direc
Leona [35]

Answer:

Fr = 26.83 [N]

Explanation:

To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.

F_{r}=\sqrt{(12)^{2}+(24)^{2}  } \\F_{r}=26.83[N]

3 0
3 years ago
A test charge of 13 mC is at a point P where an external electric field is directed to the right and has a magnitude of 4 3 106
LenKa [72]

Answer:

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C, but the direction is still to the right.

Explanation:

From coulomb's law, F = Eq

Thus,

F = E₁q₁

F = E₂q₂

Then

E₂q₂ = E₁q₁

E_2 = \frac{E_1q_1}{q_2}

where;

E₂ is the external electric field due to second test charge = ?

E₁ is the external electric field due to first test charge = 4 x 10⁶ N/C

q₁ is the first test charge = 13 mC

q₂ is the second test charge = 23 mC

Substitute in these values in the equation above and calculate E₂.

E_2 = \frac{4*10^6*13}{23} = 2.26 *10^6 \ N/C

The magnitude of the external electric field at P will reduce to 2.26 x 10⁶ N/C when 13 mC test charge is replaced with another test charge of 23 mC.

However, the direction of the external field is still to the right.

8 0
3 years ago
Does anyone know? Please help! Thank you!
Vedmedyk [2.9K]

B density will increase

7 0
3 years ago
Read 2 more answers
The magnetic flux through each turn of a 110-turn coil is given by ΦB = 9.75 ✕ 10−3 sin(ωt), where ω is the angular speed of the
Xelga [282]

Answer:

Explanation:

Given that a coil has a turns of

N = 110 turns

And the flux is given as function of t

ΦB = 9.75 ✕ 10^-3 sin(ωt),

Given that, at an instant the angular velocity is 8.70 ✕ 10² rev/min

ω = 8.70 ✕ 10² rev/min

Converting this to rad/sec

1 rev = 2πrad

Then,

ω = 8.7 × 10² × 2π / 60

ω = 91.11 rad/s

Now, we want to find the induced EMF as a function of time

EMF is given as

ε = —NdΦB/dt

ΦB = 9.75 ✕ 10^-3 sin(ωt),

dΦB/dt = 9.75 × 10^-3•ω Cos(ωt)

So,

ε = —NdΦB/dt

ε = —110 × 9.75 × 10^-3•ω Cos(ωt)

Since ω = 91.11 rad/s

ε = —110 × 9.75 × 10^-3 ×91.11 Cos(91.11t)

ε = —97.71 Cos(91.11t)

The EMF as a function of time is

ε = —97.71 Cos(91.11t)

Extra

The maximum EMF will be when Cos(91.11t) = -1

Then, maximum emf = 97.71V

8 0
3 years ago
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