What is the speed of a bobsled whose distance-time graph indicates that it traveled 114m in 30s? m/s

IrinaVladis [17]

Light year is the unit of distance. It is the distance that an object travels in one year with the speed of light.

In short, Your Answer would be "Distance"

Hope this helps!

3 0

3 years ago

A transformer consists of a 500 turn primary coil and a 2000-turnsecondary coil. If the current in the secondary is 3.0A, what i

Neporo4naja [7]

Answer:

The correct solution will be "12.0 A".

Explanation:

The given values are:

$N_p= 500 \ turn$

$N_s= 200 \ turn$

$I_s= 3.0 \ A$

By using the transformer formula, we get

⇒ $\frac{N_p}{N_s} =\frac{I_s}{I_p}$

⇒ $I_p = I_s\times \frac{N_s}{N_p}$

On substituting the given values, we get

⇒ $=3.0 \ A\times \frac{2000}{500}$

⇒ $=12.0 \ A$

8 0

3 years ago

The most common isotope of hydrogen contains a proton and an electron 'separated by about -11-27 5.0 x 10 m. The mass of proton

Brrunno [24]

Answer:

A) F_g = 4.05 10⁻⁴⁷ N, B) F_e = 9.2 10⁻⁸N, C) $\frac{F_e}{F_g}$ = 2.3 10³⁹

Explanation:

A) It is asked to find the force of attraction due to the masses of the particles

Let's use the law of universal attraction

F = $G \frac{m_1m_2}{r^2}$

let's calculate

F = $6.67 \ 10^{-11} \ \frac{9.1 \ 10^{-31} \ 1.67 \ 10 ^{-27} }{(5 \ 10^{-11})^2 }$

F_g = 4.05 10⁻⁴⁷ N

B) in this part it is asked to calculate the electric force

Let's use Coulomb's law

F = $k \ \frac{q_1q_2}{r^2}$

let's calculate

F = $9 \ 10^9 \ \frac{(1.6 \ 10^{-19} )^2}{(5 \ 10^{-11})^2}$

F_e = 9.2 10⁻⁸N

C) It is asked to find the relationship between these forces

$\frac{F_e}{F_g} = \frac{9.2 \ 10^{-8} }{4.05 \ 10^{-47} }$

= 2.3 10³⁹

therefore the electric force is much greater than the gravitational force

4 0

3 years ago

A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of

Fudgin [204]

Answer:

Part a)

$v_f = v_x = 32.77 m/s$

Part b)

T = 4.68 s

Explanation:

Part a)

Shell is fired at speed of 40 m/s at angle of 35 degree

so here we have

$v_x = 40 cos35 = 32.77 m/s$

$v_y = 40 sin35 = 22.94 m/s$

since gravity act opposite to vertical speed of the shell so at the highest point of its trajectory the vertical component of the speed will become zero

so at the highest point the speed is given

$v_f = 32.77 m/s$

Part b)

After completing the motion we know that the displacement of the object will be zero in Y direction

so we have

$\Delta y = 0$

$0 = v_y t - \frac{1}{2}gt^2$

$T = \frac{2v_y}{g}$

$T = \frac{2(22.94)}{9.81} = 4.68 s$

7 0

3 years ago

The fictional rocket ship Adventure is measured to be 65 m long by the ship's captain inside the rocket.When the rocket moves pa

kipiarov [429]

Answer:

1.04 s

Explanation:

The computation is shown below:

As we know that

t = t' × 1 ÷ (√(1 - (v/c)^2)

here

v = 0.5c

t = 1.20 -s

So,

1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)

1.20 = t' × 1 ÷ (√(1 - (0.5)^2)

1.20 = t' ÷ √0.75

1.20 = t' ÷ 0.866

t' = 0.866 × 1.20

= 1.04 s

The above formula should be applied

8 0

3 years ago