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AnnyKZ [126]
3 years ago
5

What must you learn before you ca calculate you target heart rate for exercise

Physics
1 answer:
kozerog [31]3 years ago
8 0
You must learn your resting heart rate, rate at which your heart beats during excerpts and etc you pretty much just need a heart rate monitor
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Please help me with this
amm1812

the answer might be 2.3 kilos

3 0
3 years ago
A scientist classifies some plants into two groups
gizmo_the_mogwai [7]

Answer:

I'm pretty sure the answer is C . since deciduous trees are trees that loose leaves seasonally and coniferous trees are trees that don't loose leaves seaosnally and survive through the winter.

8 0
3 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
what is the pressure exerted, what is the pressure exerted by 50kg girl as she places her weight on one shoe if the heels area i
lana [24]

Answer:

The pressure exerted by the girl is 245,000 N/m²

Explanation:

Given;

mass of the girl, m = 50 kg

area of the girl's shoe, A = 0.002 m²

The pressure exerted by the girl is calculated as follows;

P = \frac{F}{A} \\\\Where;\\F \ is \ the \ force \ exerted \ by \ girl's \ weight\\\\P = \frac{F}{A} = \frac{mg}{A} = \frac{50 \times 9.8}{0.002} = 245,000 \ N/m^2

Therefore, the pressure exerted by the girl is 245,000 N/m²

6 0
3 years ago
HELP URGENT———Why is creativity important in constructing scientific methods?
Oxana [17]

correct option is C.

Explanation:

the best to test a hypothesis is not always the most obvious way

7 0
3 years ago
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