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alekssr [168]
3 years ago
12

In a two-dimensional Cartesian system, the x-component of a vector is known, and the angle between vector and x-axis is known. W

hich operation is used to calculate the magnitude of the vector?
a. dividing by cosine
b. dividing by sine
c. multiplying by cosine
d. multiplying by sine
Physics
1 answer:
emmainna [20.7K]3 years ago
5 0
I think it’s A. Dividing by cosine
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posledela
<span>The time needed for a wave to make one complete cycle is its period.</span>
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A physical quantity, G, is defined by G = (Original mass x time)/(change in mass), what is the S.I. unit of G ?
Andrei [34K]
The gravitational constant (G) in its base SI units is

3/2
m
3
k
g
/
s
2


But is often seen written as

⋅
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⋅
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Where N is the Newton unit. N=kg ⋅
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4 0
2 years ago
if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav
allochka39001 [22]

Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

The acceleration due to gravity on the Moon = 1 ÷ 16 of that on Earth

The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

Where;

g = The acceleration due to gravity = 9.8 m/s²

Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

8 0
3 years ago
Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o
spin [16.1K]

Answer:

The true weight of the aluminium is m_{alu} = 4.5021 kg

Explanation:

Given data

m_{app} = 4.5 kg

\rho_{air} = 1.29 \frac{kg}{m^{3} }

\rho_{al} = 2.7× 10^{3} \frac{kg}{m^{3} }

The true mass of the aluminium is given by

m_{alu} = \frac{\rho_{alu}m_{app}}{\rho_{alu} -\rho_{air} }

Put all the values in above equation we get

m_{alu} = \frac{(2700)(4.5)}{2700-1.29}

m_{alu} = 4.5021 kg

Therefore the true weight of the aluminium is m_{alu} = 4.5021 kg

6 0
3 years ago
What has more kinetic energy 15 kg ball rolling north at 15 m/s or a 15 kg ball rolling backwards at 7m/s
Setler79 [48]

Answer:

15 kg ball

Explanation:

6 0
3 years ago
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