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Lesechka [4]
3 years ago
13

What is the SI unit of acceleration and work?​

Physics
1 answer:
Lerok [7]3 years ago
6 0
A=m/s^2(meter per second square)
Work=joule
You might be interested in
Can you find the magnetic field based on force? a straight segment of wire 35.0 cm long carrying a current of 1.40 a is in a uni
Lostsunrise [7]
The force exerted by a magnetic field on a wire carrying current is:
F=ILB \sin \theta
where I is the current, L the length of the wire, B the magnetic field intensity, and \theta the angle between the wire and the direction of B.

In our problem, the force is F=0.20 N. The current is I=1.40 A, while the length of the wire is L=35.0 cm=0.35 m. The angle between the wire and the magnetic field is 53 ^{\circ}, so we can re-arrange the formula and substitute the numbers to find B:
B= \frac{F}{IL \sin \theta}= \frac{0.20 N}{(1.40 A)(0.35 m)(\sin 53^{\circ})}=0.51 T
3 0
3 years ago
Solve for M₂
soldi70 [24.7K]

Explanation:

M₂ = Fr²/GM₁

M₂ = [(132N)(.243m)²]/[(6.67*10^-11N*m²/kg)(1.175*10^4kg)]

M₂ = (7.79N*m²)/(7.84*10^-7N*m²)

M₂ = 9.94*10^6 kg

5 0
3 years ago
A cell membrane consists of an inner and outer wall separated by a distance of approximately 10nm. Assume that the walls act lik
Nina [5.8K]

Answer:

1 × 10⁶ N/C

Explanation:

The magnitude of the electric field between the membrane = surface density / permittivity of free space = 10 ⁻⁵C/ m² / (8.85 × 10⁻¹²N⁻¹m⁻²C²) = 1.13 × 10⁶ N/C approx 1 × 10⁶ N/C

4 0
3 years ago
A flat coil of wire consisting of 20 turns, each with an area of 50 cm2, is positioned perpendicularly to a uniform magnetic fie
jenyasd209 [6]

Answer:

0.5 A

Explanation:

N = 20, A = 50 cm^2 = 50 x 10^-4 m^2, dB = 6 - 2 = 4 T, dt = 2 s, R = 0.4 ohm

The induced emf is given by

e = - N dФ/dt

Where, dФ/dt is the rate of change of magnetic flux.

Ф = B A

dФ/dt = A dB/dt

so,

e = 20 x 50 x 10^-4 x 4 / 2 = 0.2 V

negative sign shows the direction of magnetic field.

induced current, i = induced emf / resistance = 0.2 / 0.4 = 0.5 A

5 0
3 years ago
A low C (f=65Hz) is sounded on a piano. If the length of the piano wire is 2.0 m and
WITCHER [35]

Answer:

T = 676 N

Explanation:

Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g/m^{2} = 0.005 kg

A stationary wave that is set up in the string has a frequency of;

f = \frac{1}{2L}\sqrt{\frac{T}{M} }

⇒      T = 4L^{2}f^{2}M

Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.

But M = L × ρ = (2 × 0.005) = 0.01 kg/m

T = 4 × 2^{2} ×65^{2} × 0.01

   = 4 × 4 ×4225 × 0.01

   = 676 N

Tension of the wire is 676 N.

4 0
3 years ago
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