HELP PLS MARKING BRANLIST 100 Pts TAKING TEST RN

What is the length of the hypotenuse of a right triangle, whose legs have lengths of 12 meters and 35 meters?

Eduardwww [97]

Answer:

c

Explanation:

6 0

3 years ago

a sensor light installed on the edge of a home can detect motion for a distance of 50 feet in front and with a range of motion o

Dvinal [7]

Answer:

4363.3231 feets²

Explanation:

Given that :

Distance, r = 50 ft

θ = 200°

The arc length of area covered :

Arc length = θ/360° * πr²

Arc length = (200/360) * 50 ft ^2 * π

Arc length = 0.5555555 * 2500 * π

Arc length = 4363.3231 feets²

7 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago

An object with a mass of 1.5 kg changes its velocity +15 m/s during a time interval of 3.5 seconds what impulse was delivered to

Arada [10]

The answer is 10.5 kg m/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So:
a = Δv/Δt ⇒ F = m · a = m · Δv/Δt
F = m · Δv/Δt ⇒ I = m · Δv/Δt · Δt
Since Δt can be cancelled out, impulse can be expressed as:
I = m · Δv = m · (v2 - v1)

It is given:
m = 1.5 kg
v1 = 15 m/s
v2 = 22 m/s

I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.

7 0

4 years ago

I need help. <br> How do I fill this out?

enyata [817]

Answer:

maybe try searching it up

6 0

3 years ago

Read 2 more answers