If a projectile is launched at a velocity of 20 m/s, is it in equilibrium?<br> Why or why not?

A block of mass m = 1.5 kg is released from rest at a height of H = 0.81 m on a curved frictionless ramp. At the foot of the ram

kogti [31]

Answer:

0.31 m

Explanation:

m = mass of the block = 1.5 kg

H = height from which the block is released on ramp = 0.81 m

k = spring constant of the spring = 250 N/m

x = maximum compression of the spring

using conservation of energy

Spring potential energy gained by spring = Potential energy lost by block

(0.5) k x² = mgH

(0.5) (250) x² = (1.5) (9.8) (0.81)

x = 0.31 m

7 0

3 years ago

As we move ACROSS the periodic table in a row, the electronegativity increases decreases

trasher [3.6K]

It increases across a period but it decreases down a group.

6 0

3 years ago

Read 2 more answers

A uniform, solid sphere of radius 2.50 cm and mass 4.75 kg starts with a purely translational speed of 3.00 m/s at the top of an

allsm [11]

Answer:

The final translational seed at the bottom of the ramp is approximately 4.84 m/s

Explanation:

The given parameters are;

The radius of the sphere, R = 2.50 cm

The mass of the sphere, m = 4.75 kg

The translational speed at the top of the inclined plane, v = 3.00 m/s

The length of the inclined plane, l = 2.75 m

The angle at which the plane is tilted, θ = 22.0°

We have;

$K_i$ + $U_i$ = $K_f$ + $U_f$

K = (1/2)×m×v²×(1 + I/(m·r²))

I = (2/5)·m·r²

K = (1/2)×m×v²×(1 + 2/5) = 7/10 × m×v²

U = m·g·h

h = l×sin(θ)

h = 2.75×sin(22.0°)

∴ 7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) = 7/10 × 4.75× $v_f$ ² + 0

7/10×4.75×3.00² + 4.75×9.81×2.75×sin(22.0°) ≈ 77.93

∴ 77.93 ≈ 7/10 × 4.75× $v_f$ ²

$v_f$ ² = 77.93/(7/10 × 4.75)

$v_f$ ≈ √(77.93/(7/10 × 4.75)) ≈ 4.84

The final translational seed at the bottom of the ramp, $v_f$ ≈ 4.84 m/s.

3 0

2 years ago

The common value for the speed of light in a vacuum is:

OLEGan [10]

The common value for “Speed of light in vacuum” is $\bold{3 \times 10^{8}}$ metre per second.

Answer: Option b

<u>Solution: </u>

Speed of light can be defined as the speed with which light waves propagate in different medium. In vacuum, speed of light is 186,282 miles per second or 299,792 km/s which is rounded off as $3 \times 10^{8} \mathrm{m} / \mathrm{s}$ .

“Speed of light in vacuum” is a universal constant and usually represented by ‘c’. Light waves travels at a speed of $3 \times 10^{8}$ metre per second in vacuum.

5 0

3 years ago

A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N

Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

$w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s$

Force on object is

$F(t)=-mAw^{2}Cos(wt)$

Substitute given values

So

$F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N$

So

Magnitude F(t)=26.6 N

Direction: -x

4 0

3 years ago

If a projectile is launched at a velocity of 20 m/s, is it in equilibrium? Why or why not?