The equation most often associated with Newton's Second Law of Motion is<br> Answer here

a. As you coast down a hill on your bicycle, you accelerate at 0.5 m/s2. If the total mass of your body and the bicycle is 80 ki

zalisa [80]

According to Newton’s second law of motion:
F=ma
=80*0.5
=40N

7 0

3 years ago

Calculate the mass 9f the earth, assuring that uts is sphere with radius 6.67×10^6m.

MA_775_DIABLO [31]

Answer:

6.86 × 10²⁴ kg

Explanation:

The mass of the earth m = density of earth, ρ × volume of earth, V

m = ρV

The density of the earth, ρ = 5515 kg/m³ and since the earth is a sphere, its volume is the volume of a sphere V = 4πr³/3 where r = radius of the earth = 6.67 × 10⁶ m

Since m = ρV

m = ρ4πr³/3

So, substituting the values of the variables into the equation for the mass of the earth, m, we have

m = 5515 kg/m³ × 4π(6.67 × 10⁶ m)³/3

m = 5515 kg/m³ × 4π × 296.741 × 10¹⁸ m³/3

m = 5515 kg/m³ × 1189.9639π × 10¹⁸ m³/3

m = 6546105.64378π × 10¹⁸ kg/3

m = 20565197.400122 × 10¹⁸ kg/3

m = 6855065.8 × 10¹⁸ kg

m = 6.8550658 × 10²⁴ kg

m ≅ 6.86 × 10²⁴ kg

8 0

3 years ago

Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment

Drupady [299]

Answer:

$u_K=0.862$

Explanation:

The force of friction between the quails feet and the ground is:

$F=m*a$

$F_K=m*a$

$F_K=u_k*m*g$

$u_K*m*g=m*a_c$

$u_K*g=a$

$u_K=\frac{a_c}{g}$

$a_c=\frac{v^2}{r}$

So the coefficient of static is solve

$u_K=\frac{\frac{v^2}{r}}{g}$

$u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}$

$u_K=0.862$

4 0

3 years ago

Electrons and protons travel from the Sun to the Earth at a typical velocity of 3.83 ✕ 105 m/s in the positive x-direction. Thou

Leona [35]

Answer:

$F=2.84*10^{-26}N$ & -y direction

$F=2.84*10^{-26}N$ & +y direction

Explanation:

From the question we are told that:

Speed of electron $V_e=3.83 * 10^5 m/s$ +x direction

Earths magnetic field $B_e=3.04 * 10^-^8$ +z direction

a)

Generally the equation for magnetic force $F_m$ is mathematically given by

$F=q(V_e*B_e)$

where

$q=1.6*10^{-19}c\\\=i*\=z=-\=j$

$F=1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$F=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ -y direction

b)

Generally the equation for magnitude and direction of the magnetic force on an electron. is mathematically given by

$\=F'=-1.6*10^{-19}(3.83 * 10^5 m/s*3.04 * 10^-^8)$

$\=F'=-2.84*10^{-26}N \=j$

Magnitude & Direction

$F=2.84*10^{-26}N$ & +y direction

5 0

3 years ago

the temperature of a 2.0-kg increases by 5*c when 2,000 J of thermal energy are added to the block. What is the specific heat of

nata0808 [166]

To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
c = 2000/ 10
c = 200

Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1

3 0

3 years ago

The equation most often associated with Newton's Second Law of Motion is Answer here