Which of the following occurs in meiosis that does not occur in mitosis

A parcel has a mass of 500 g .calculate its weight (assume the gravitational field strength is 10 N/Kg

Effectus [21]

Answer:

5N

Explanation

convert grams to kg and multipy with 10 (.5 *10)=5N

5 0

3 years ago

True or false. A convection current is the circular motion of a fluid caused by the rising of heated fluid

xz_007 [3.2K]

True that is what a convection current is

7 0

2 years ago

A certain quantity of steam has a temperature of 100.0 oC. To convert this steam into ice at 0.0 oC, energy in the form of heat

KonstantinChe [14]

Answer:

2452.79432 m/s

Explanation:

m = Mass of ice

$L_s$ = Latent heat of steam

$s_w$ = Specific heat of water

$L_i$ = Latent heat of ice

v = Velocity of ice

$\Delta T$ = Change in temperature

Amount of heat required for steam

$Q_1=mL_s\\\Rightarrow Q_1=m(2.256\times 10^6)$

Heat released from water at 100 °C

$Q_2=ms_w\Delta T\\\Rightarrow Q_2=m4186\times (100-0)\\\Rightarrow Q_2=m0.4186\times 10^6$

Heat released from water at 0 °C

$Q_3=mL_i\\\Rightarrow Q_3=m(333.5\times 10^3)\\\Rightarrow Q_3=m(0.3335\times 10^6)$

Total heat released is

$Q=Q_1+Q_2+Q_3\\\Rightarrow Q=m(2.256\times 10^6)+m0.4186\times 10^6+m(0.3335\times 10^6)\\\Rightarrow Q=3008100m$

The kinetic energy of the bullet will balance the heat

$K=Q\\\Rightarrow \frac{1}{2}mv^2=3008100m\\\Rightarrow v=\sqrt{2\times 3008100}\\\Rightarrow v=2452.79432\ m/s$

The velocity of the ice would be 2452.79432 m/s

6 0

3 years ago

A block weighting 400kg rests on a horizontal surface and support on top of it another block of weight 100kg placed on the top o

masha68 [24]

The horizontal force applied to the block is approximately 1,420.84 N

The known parameters;

The mass of the block, w₁ = 400 kg

The orientation of the surface on which the block rest, w₁ = Horizontal

The mass of the block placed on top of the 400 kg block, w₂ = 100 kg

The length of the string to which the block w₂ is attached, l = 6 m

The coefficient of friction between the surface, μ = 0.25

The state of the system of blocks and applied force = Equilibrium

Strategy;

Calculate the forces acting on the blocks and string

The weight of the block, W₁ = 400 kg × 9.81 m/s² = 3,924 N

The weight of the block, W₂ = 100 kg × 9.81 m/s² = 981 N

Let <em>T</em> represent the tension in the string

The upward force from the string = T × sin(θ)

sin(θ) = √(6² - 5²)/6

Therefore;

The upward force from the string = T×√(6² - 5²)/6

The frictional force = (W₂ - The upward force from the string) × μ

The frictional force, $F_{f2}$ = (981 - T×√(6² - 5²)/6) × 0.25

The tension in the string, T = $F_{f2}$ × cos(θ)

∴ T = (981 - T×√(6² - 5²)/6) × 0.25 × 5/6

Solving, we get;

$T = \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \approx 183.27$

$Frictional \ force, F_{f2} = \left (981 - \dfrac{5886}{\sqrt{6^2 - 5^2} + 28.8} \times \dfrac{\sqrt{6^2 - 5^2} }{6} \times 0.25 \right) \approx 219.92$

The frictional force on the block W₂, $F_{f2}$ ≈ 219.92 N

Therefore;

The force acting the block w₁, due to w₂ $F_{w2}$ = 219.92/0.25 ≈ 879.68

The total normal force acting on the ground, N = W₁ + $\mathbf{F_{w2}}$

The frictional force from the ground, $\mathbf{F_{f1}}$ = N×μ + $\mathbf{F_{f2}}$ = P

Where;

P = The horizontal force applied to the block

P = (W₁ + $\mathbf{F_{w2}}$ ) × μ + $\mathbf{F_{f2}}$

Therefore;

P = (3,924 + 879.68) × 0.25 + 219.92 ≈ 1,420.84

The horizontal force applied to the block, P ≈ 1,420.84 N

Learn more about friction force here;

brainly.com/question/18038995

3 0

2 years ago

Consider a thick cylindrical wire of radius a. What must be the functional form of the current density J so that the magnetic fi

NikAS [45]

Answer:

Explanation:

Check the attachment

3 0

3 years ago