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4 years ago

Imagine a glass box that is completely sealed and in that box is a cactus in a small pot. Which statement is most accurate?

1 answer:

Natasha2012 [34]4 years ago

3 0

The Answer Is B (: because if it was closed no light could get throw and its not d because air doesnt circulate throw water xD HOPE THIS HELPS ANSWER IS REALLY B ♥

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A motorcyclist drives from a to b with the uniform speed of 30 km/h-1 and returns back with the speed of 20 km/h-1.find the aver

maxonik [38]

30+20 =50
For average 50/2=25

4 0

3 years ago

Read 2 more answers

If a = 8i + j - 2k and b = 5i - 3j + k show that a) a x b = -5i - 18j - 29k b) b X a = 50 + 18j +29k

loris [4]

Recall the definition of the cross product with respect to the unit vectors:

i × i = j × j = k × k = 0

i × j = k

j × k = i

k × i = j

and that the product is anticommutative, so that for any two vectors u and v, we have u × v = - (v × u). (This essentially takes care of part (b).)

Now, given a = 8i + j - 2k and b = 5i - 3j + k, we have

a × b = (8i + j - 2k) × (5i - 3j + k)

a × b = 40 (i × i) + 5 (j × i) - 10 (k × i)

… … … … - 24 (i × j) - 3 (j × j) + 6 (k × j)

… … … … + 8 (i × k) + (j × k) - 2 (k × k)

a × b = - 5 (i × j) - 10 (k × i) - 24 (i × j) - 6 (j × k) - 8 (k × i) + (j × k)

a × b = - 5k - 10j - 24k - 6i - 8j + i

a × b = -5i - 18j - 29k

7 0

3 years ago

A mass of 0.273 kg is placed on an incline of 38.382 degrees. It is attached by a rope over a pulley to a mass of 0.254 kg. Find

Natali5045456 [20]

here tension in the string is counter balanced by weight of block of mass m1

so we can say

$T = m_1g$

$T = 0.254 \times 9.8 = 2.49 N$

now on the other side the block which is placed on the inclined plane

we can say that component of weight of the block and friction force is counter balanced by tension force

$m_2g sin\theta + F_f = T$

now we can plug in all values to find the friction force

$0.273 \times 9.8 sin38.382 + F_f = 2.49$

$1.66 + F_f = 2.49$

$F_f = 0.83 N$

so it will have 0.83 N force on it due to friction

now to find the friction coefficient

$F_f = \mu \times F_n$

here we know that

$F_n = m_2gcos\theta$

$F_n = 0.273 \times 9.8 \times cos38.382 = 2.1 N$

now from above equation

$0.83 = \mu \times 2.1$

$\mu = 0.38$

so friction coefficient will be 0.38

8 0

3 years ago

If 61.4 cm of copper wire (diameter = 1.08 mm, resistivity = 1.69 × 10-8Ω·m) is formed into a circular loop and placed perpendic

harkovskaia [24]

Answer:

the thermal energy generated in the loop = $6.64*10^{-4} \ W$

Explanation:

Given that;

The length of the copper wire L = 0.614 m

Radius of the loop r = $\frac{L}{2 \pi}$

r = $\frac{0.614}{2 \pi}$

r = 0.0977 m

However , the area of the loop is :

$A_L = \pi r^2$

$A_L = \pi (0.0977)^2$

$A_L = 0.02999 \ m^2$

Change in the magnetic field is $\frac{dB}{dt}= 0.0914 \ T/s$

Then the induced emf e = $A_L \frac{dB}{dt}$

e = $0.02999 * 0.0914$

e = 2.74 × 10⁻³ V

resistivity of the copper wire $\rho = 1.69* 10^{-8}$ Ω m

diameter of the wire = 1.08 mm

radius of the wire = 0.54 mm = 0.54 × 10⁻³ m

Thus, the resistance of the wire R = $\frac {\rho L}{\pi r^2}$

R = $\frac{(1.69*10^{-8})(0.614)}{ \pi (0.54*10^{-3})^2}$

R = 1.13× 10⁻² Ω

Finally, the thermal energy generated in the loop (i.e the power) = $\frac{e^2}{R}$

= $\frac{(2.74*10^{-3})^2}{1.13*10^{-2}}$

= $6.64*10^{-4} \ W$

8 0

3 years ago

34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then

velikii [3]

Answer:

$F_H_n=230.04 N$

The Required horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

$F_H_n=F_f\\F_h=mg*u$

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

$200=mg*u$ Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

$F_H=(1.42m*g*.81u)$ Eq (2)

Divide Eq(2) and Eq (1)

$\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N$

The Required horizontal force is 230.04N

4 0

3 years ago