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olga nikolaevna [1]
3 years ago
9

If a melon has a a mass of 1 kg, how much does the melon weigh?

Physics
2 answers:
ipn [44]3 years ago
5 0
1 kg is approximately 10N

Hope this helps! Please mark brainliest :)
CaHeK987 [17]3 years ago
3 0

Answer:

A. 10 N

Explanation:

the weight of melon is 1×10=10N

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3 years ago
Which descriptions apply to Mendels pea plant experiments ? Select three options
const2013 [10]

used cross breeding to purposely breed plan

studied a variety of pea plant traits

studied several generations of plants

In order to breed or cross one plant with another, Mendel opened the petals and removed the anthers from the flower and dusted the pistil with pollen from the plant he wished to cross it with and covered the flower with a small bag to prevent the pollen in the air from landing in the pistil. This process is called cross pollination.

Mendel considered 7 characters of pea plant for his study and did the study for several generations.

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How many different values of ml are possible in the 5d sublevel?
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4 0
4 years ago
A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
This exercise uses the radioactive decay model. After 3 days a sample of radon-222 has decayed to 58% of its original amount. (a
Vadim26 [7]

Answer: a) 3.85 days

b) 10.54 days

Explanation:-

Expression for rate law for first order kinetics is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant  = ?

t = time taken for decomposition  = 3 days

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{58}{100}\times 100=58g

First we have to calculate the rate constant, we use the formula :

Now put all the given values in above equation, we get

k=\frac{2.303}{3}\log\frac{100}{58}

k=0.18days^{-1}

a) Half-life of radon-222:

t_{\frac{1}{2}}=\frac{0.693}{k}

t_{\frac{1}{2}}=\frac{0.693}{0.18}=3.85days

Thus half-life of radon-222 is 3.85 days.

b) Time taken for the sample to decay to 15% of its original amount:

where,

k = rate constant  = 0.18days^{-1}

t = time taken for decomposition  = ?

a = let initial amount of the reactant  = 100 g

a - x = amount left after decay process  = \frac{15}{100}\times 100=15g

t=\frac{2.303}{0.18}\log\frac{100}{15}

t=10.54days

Thus it will take 10.54 days for the sample to decay to 15% of its original amount.

3 0
3 years ago
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