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3 years ago

What is the momentum of a 52 kg carton that slides at 5.0m/s across an icy surface?

Physics

1 answer:

nika2105 [10]3 years ago

5 0

Formula for momentum is mass times velocity so you do 52 kg times 5 m/s and that equals 260 kg x m/s

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. The inner and outer surfaces of a 4-m × 7-m brick wall of thickness 30 cm and thermal conductivity 0.69 W/m-K are maintained a

Anni [7]

Answer:

$\frac{dQ}{dt} = 966 W$

Explanation:

As we know that the rate of heat transfer due to temperature difference is given by the formula

$\frac{dQ}{dt} = \frac{KA(\Delta T)}{L}$

here we know that

$K = 0.69 W/m-K$

A = 4 m x 7 m

thickness = 30 cm

temperature difference is given as

$\Delta T = 20 - 5 = 15 ^oC$

now we have

$\frac{dQ}{dt} = \frac{(0.69W/m-K)(28 m^2)(15)}{0.30}$

$\frac{dQ}{dt} = 966 W$

4 0

3 years ago

If the speed of sound is approximately 671 mi/hr, about how many meters per second is it? (1mile=1609m)

zzz [600]

671mi/hr

= 671/60min (calculates miles/min)
= (671/60) ÷ 60seconds (calculates miles/sec)

((671/60) ÷ 60)× 1609m
= 299.899 meters/sec
= 299.90m (round off to 2 decimals )

3 0

3 years ago

If it takes 150N of force to move a box 10 meters. what is the work done on the box?

sergeinik [125]

Answer:

1500 Joules

Explanation:

Work = Force x Distance

When multiplying by 10 you simply shift all the digits to the

left and append a 0 to the end.

so 150 x 10 = 1500 Joules

3 0

3 years ago

Read 2 more answers

A single circular loop of wire of radius 0.45 m carries a constant current of 2.4 A. The loop may be rotated about an axis that

Snowcat [4.5K]

Answer:

The magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T

Explanation:

Given;

radius of the wire, r = 0.45 m

current on the loop, I = 2.4 A

angle of inclination, θ = 36⁰

torque on the coil, τ = 1.5 N.m

The torque on the coil is given by;

τ = NIBAsinθ

where;

B is the magnetic field

Area of the loop is given by;

A = πr² = π(0.45)² = 0.636 m

τ = NIBAsinθ

1.5 = (1 x 2.4 x 0.636 x sin36)B

1.5 = 0.8972B

B = 1.5 / 0.8972

B = 1.67 T

Therefore, the magnitude of the uniform magnetic field exerting this torque on the loop is 1.67 T

8 0

3 years ago

3. A certain horizontal east-west lined wire has a mass of 0.2kg per meter of length and carries a current I. Impressed on the w

erica [24]

Answer:

i = 4.9 A

Explanation:

The expression for the magnetic force in a wire carrying a current is

F = i L x B

bold letters indicate vectors.

The direction of the cable is towards the East, the direction of the magnetic field is towards the North, so the vector product is in the vertical direction (z-axis) upwards and the weight of the cable is vertical downwards. Let's apply the equilibrium condition

F - W = 0

i L B = m g

They indicate the linear density of the cable λ = 0.2 kg / m

λ = m / L

m = λ L

we substitute

i B = λ g

i = $\frac{ \lambda \ g}{B}$

let's calculate

i = 0.2 9.8 / 0.4

i = 4.9 A

3 0

3 years ago