Press the<br>story 14th<br>the won (Takedans far 1299/

Igoryamba

I would say d if i was you

7 0

3 years ago

Calculate the displacement in m and velocity in m/s at the following times for a rock thrown straight down with an initial veloc

svet-max [94.6K]

Incomplete question as time is missing.I have assumed some times here.The complete question is here

Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, (d) 2.00 s, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 10 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Explanation:

Given data

Vi=10 m/s

S=70 m

(a) t₁=0.5 s

(b) t₂=1 s

(c) t₃=1.5 s

(d) t₄=2 s

(e) t₅=2.5 s

To find

Displacement S from t₁ to t₅

Velocity V from t₁ to t₅

Solution

According to kinematic equation of motion and given information conclude that v is given by

$v=v_{i}+gt\\$

Also get the equation of displacement

$S=v_{i}t+(1/2)gt^{2}$

These two formula are used to find velocity as well as displacement for time t₁ to t₅

For t₁=0.5 s

$v_{1}=v_{i}+gt\\v_{1}=(10m/s)+(9.8m/s^{2} ) (0.5s)\\v_{1}=14.9m/s\\ And\\S_{1} =v_{i}t+(1/2)gt^{2}\\ S_{1}=(10m/s)(0.5s)+(1/2)(9.8m/s^{2} )(0.5s)^{2} \\S_{1}=6.225m$

For t₂

$v_{2}=v_{i}+gt\\v_{2}=(10m/s)+(9.8m/s^{2} ) (1s)\\v_{2}=19.8m/s\\ And\\S_{2} =v_{i}t+(1/2)gt^{2}\\ S_{2}=(10m/s)(1s)+(1/2)(9.8m/s^{2} )(1s)^{2} \\S_{2}=14.9m$

For t₃

$v_{3}=v_{i}+gt\\v_{3}=(10m/s)+(9.8m/s^{2} ) (1.5s)\\v_{3}=24.7m/s\\ And\\S_{3} =v_{i}t+(1/2)gt^{2}\\ S_{3}=(10m/s)(1.5s)+(1/2)(9.8m/s^{2} )(1.5s)^{2} \\S_{3}=26.025m$

For t₄

$v_{4}=v_{i}+gt\\v_{4}=(10m/s)+(9.8m/s^{2} ) (2s)\\v_{4}=29.6m/s\\ And\\S_{4} =v_{i}t+(1/2)gt^{2}\\ S_{4}=(10m/s)(2s)+(1/2)(9.8m/s^{2} )(2s)^{2} \\S_{4}=39.6m$

For t₅

$v_{5}=v_{i}+gt\\v_{5}=(10m/s)+(9.8m/s^{2} ) (2.5s)\\v_{5}=34.5m/s\\ And\\S_{5} =v_{i}t+(1/2)gt^{2}\\ S_{5}=(10m/s)(2.5s)+(1/2)(9.8m/s^{2} )(2.5s)^{2} \\S_{5}=55.625m$

4 0

4 years ago

A snowboarder starts from rest at the top of a slope. They accelerate to 35 m/s over 9 seconds. What is their acceleration?

leonid [27]

a=v-u/t

v=35m/s

t=9

a=35/9
a=3.88m/s²

4 0

3 years ago

Por favor, necesito ayuda urgente 4. vectores perpendiculares de 25 y 40 unidades cada uno. Hallar gráfica y numericamente el ve

Darina [25.2K]

Answer:

4) R = 47.17 units , 5) R= 10,29 unidades, 6) R= 2994,4 km , θ = -33,7

Explanation:

Este es un ejercicio de adición de vectores

4) como los vectores son perpendiculares.

Para encontrar la resultante podemos usar el teorema de pitoras

R = √ a² + b²

R = √√ ( 25² + 40²)

R = 47,17 unidades

5) Este caso como el angulo es diferente de 90 debemos usar la relación de pitoras completa

R² = a² + b² + 2 a b cos θ

donde el angulo es entre los vectores a y b

R² = 12² + 16² – 2 12 16 cos 40

R²= 400 – 294,16

R= 10,29 unidades

6) En este caso los dos desplazamientos son perpendiculares, por lo cual Usamos el teorema de Pitágoras

R = √ (2400² + 1600²)

R= 2994,4 km

para el angulo de este desplazamiento usamos trigonometría

tan θ = y/x

θ = tan⁻¹ y/x

θ = tan⁻¹ 1600/(-2400)

θ = -33,7

TRASLATE

This is a vector addition exercise

4) as the vectors are perpendicular.

To find the result we can use the Poreor theorem

R = √ a² + b²

R = √ (25² + 40²)

R = 47.17 units

5) This case, as the angle is different from 90, we must use the complete ratio of the pitoras

R² = a² + b² + 2 a b cos θ

where the angle is between vectors a and b

R² = 12² + 16² - 2 12 16 cos 40

R² = 400 - 294.16

R = 10.29 units

6) In this case the two displacements are perpendicular, which is why we use the Pythagorean theorem

R = √ (2400² + 1600²)

R = 2,994.4 km

for the angle of this displacement we use trigonometry

tan θ = y / x

θ = tan⁻¹ y / x

θ = tan⁻¹ 1600 / (- 2400)

θ = -33.7

3 0

3 years ago

Which mathematically describes the wave properties of electrons?.

Afina-wow [57]

The Quantum Theory

Hope this helps

8 0

3 years ago

Read 2 more answers

Press thestory 14ththe won (Takedans far 1299/​

Press thestory 14ththe won (Takedans far 1299/