What Type of Wave is Shown Below? PLZ ANSWER I'M ALREADY FAILING AT THIS SUBJECT!

The slope of the line tangent to the curve on a position-time graph at a specific time is the

Rudik [331]

Answer:

I do I make a brinliest can you please can me

7 0

3 years ago

A Parachutist with a camera, with descending at a speed of 12.5m/s, releases, the camera at an altitude of 64.3m. What is the ma

jonny [76]

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

$h = ut+\dfrac{gt^2}{2}$

Putting all given values, we get :

$12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3$

t = 2.56 and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

7 0

3 years ago

A car whose total mass is 800kg travelling with a uniform velocity of 20m/s suddenly observes a stationary dog 50m ahead on its

maxonik [38]

Answer:

The driver hits the stationery dog because the applied force is less than required force

Explanation:

Kinetic energy will be given by

$KE=0.5mv^{2}$ where m is the mass of the vehicle and v is the speed/velocity of the vehicle.

Substituting 800 Kg for m and 20 m/s for v we obtain

$KE=0.5*800*(20 m/s)^{2}=160,000$

Frictional force by vehicle pads is given by

$Fr=\frac {KE}{d}$ where d is the distance moved

Substituting 160000 for KE and 50 m for d we obtain

$Fr=\frac {160000}{50}=3200 N$

Therefore, the vehicle hits the dog since the required force is 3200N but the driver applied only 2000 N

7 0

2 years ago

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Rainbow [258]

Answer:

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Explanation:

6 0

2 years ago

Read 2 more answers

A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force

djyliett [7]

Answer:

a) U = 0 J

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0

3 years ago