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3 years ago

A 25kg box fell 200m with an acceleration of 5 m/s2. with what force did it hit the floor when it landed?

1 answer:

6 0

According to Newton's Second Law of motion, the net force acting on the object is equal to its mass multiplied by its acceleration. In formula, it is written as

Net Force =mass * acceleration
Net force = 25 kg * 5m/s^2
Net force = 125 Newtons

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What is a non example of linear relationship

xxTIMURxx [149]

Neither the speed nor the distance of a falling object is linearly related to time.

4 0

3 years ago

An unknown substance has a mass of 11.9 g . When the substance absorbs 1.071×102 J of heat, the temperature of the substance is

Mashcka [7]

Answer: The most likely identity of the substance is iron.

Explanation:

The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.

$Q=m\times c\times \Delta T$

Q = Heat absorbed= $1.071\times 10^2$ Joules

m= mass of substance = 11.9 g

c = specific heat capacity = ?

Initial temperature of the water = $T_i$ = 25.0°C

Final temperature of the water = $T_f$ = 45.0°CChange in temperature , $\Delta T=T_f-T_i=(45-25)^0C=20^0C$

Putting in the values, we get:

$1.071\times 10^2=11.9\times c\times 20^0C$

$c=0.45J/g^0C$

The specific heat of 0.45 is for iron and thus the substance is iron.

3 0

3 years ago

A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

I NEED THIS A SOON AS POSSIBLE

vivado [14]

Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.

6 0

2 years ago

Read 2 more answers

A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed

galina1969 [7]

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

3 0

3 years ago