If you travel 450 meters in 40 seconds, what is your average speed in meters per

We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

Arisa [49]

Answer:

$31.6\:\mathrm{m/s}$

Explanation:

The elastic potential energy of a spring is given by $Us=\frac{1}{2}kx^2$ , where $k$ is the spring constant of the spring and $x$ is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by $KE=\frac{1}{2}mv^2$ , where $m$ is the mass of the object and $v$ is the object's velocity.

Thus, we have:

$Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2$

Substituting given values, we get:

$\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}$

4 0

3 years ago

(a) A space vehicle is launched vertically upward from the Earth's surface with an initial speed of vi that is comparable to but

ad-work [718]

Energy Conservation Theory,

$(k+v)_i=(k+v)_f \quad \text { (No air resistone)}\\$

$\frac{1}{2} m v_i^2-\frac{G m M_\epsilon}{R_\epsilon}=0-\frac{G m M_\epsilon}{R_\epsilon+h}$

$v_{e x^2}{ }^2-v_i^2=\frac{v_{e^2 R_t} R_t}{R_t t h}\\&\frac{1}{v_{B C^2-v_1^2}^2}=\frac{R_E+h}{v_{e^2 R_E} R_E}\\\\\\h=\frac{R_E V_1^2}{v_{\text {esc }}^2-v_1^{\beta^2}}$

<h3>What is law of energy conservation?</h3>

The principle of energy conservation states that energy is neither created nor destroyed. It may change from one sort to another. Just like the mass conservation rule, the legitimacy of the preservation of energy depends on experimental perceptions; hence, it is an experimental law. The law of preservation of energy, too known as the primary law of thermodynamics

To learn more about Energy Conservation Theory, visit;

brainly.com/question/8004680

#SPJ4

7 0

2 years ago

A steel ball of mass 0.500 kg is fastened to a cord that is 70.0 cm long and fixed at the far end. The ball is then released whe

Liula [17]

Answer:

a) v₁fin = 3.7059 m/s (→)

b) v₂fin = 1.0588 m/s (→)

Explanation:

a) Given

m₁ = 0.5 Kg

L = 70 cm = 0.7 m

v₁in = 0 m/s ⇒ Kin = 0 J

v₁fin = ?

hin = L = 0.7 m

hfin = 0 m ⇒ Ufin = 0 J

The speed of the ball before the collision can be obtained as follows

Einitial = Efinal

⇒ Kin + Uin = Kfin + Ufin

⇒ 0 + m*g*hin = 0.5*m*v₁fin² + 0

⇒ v₁fin = √(2*g*hin) = √(2*(9.81 m/s²)*(0.70 m))

⇒ v₁fin = 3.7059 m/s (→)

b) Given

m₁ = 0.5 Kg

m₂ = 3.0 Kg

v₁ = 3.7059 m/s (→)

v₂ = 0 m/s

v₂fin = ?

The speed of the block just after the collision can be obtained using the equation

v₂fin = 2*m₁*v₁ / (m₁ + m₂)

⇒ v₂fin = (2*0.5 Kg*3.7059 m/s) / (0.5 Kg + 3.0 Kg)

⇒ v₂fin = 1.0588 m/s (→)

7 0

4 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

Drupady [299]

Answer:

$4.5\times 10^{-5} T$

Explanation:

We are given that

Current in wire=40 A

Magnetic field= $B_1=3.5\times 10^{-5}$ T( vertically downward)

We have to find the resultant magnitude of the magnetic field 29 cm above the wire and 29 cm below the wire.

According to Bio-Savart law, the magnetic field exerted by the wire at distance R is given by

$B_{wire}=B_2=\frac{\mu_0I}{2\pi R}$

We have R=29 cm= $\frac{29}{100}=0.29 m$

1 m=100 cm

Substitute the values in the given formula

$B_2=\frac{4\pi\times 10^{-7}\times 40}{2\times \pi\times 0.29}=\frac{2\times 40\times 10^{-7}}{0.29}=2.76\times 10^{-5} T$

The resultant magnetic field is given by

$B=\sqrt{B^2_1+B^2_2}$

Substitute the values then we get

$B=\sqrt{(3.5\times 10^{-5})^2+(2.76\times 10^{-5})^2}$

$B=4.5\times 10^{-5} T$

The resultant magnitude of magnetic field is same above and below the wire as it is at same distance.

The resultant magnitude of the magnetic field 29 cm below the wire= $4.5\times 10^{-5} T$

Hence, the resultant magnitude of the magnetic field 29 cm above the wire= $4.5\times 10^{-5} T$

7 0

3 years ago

A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?

iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ 60 kg (Ans)

Hope it helps you

6 0

3 years ago