Ask question

Ask question

All categories

3 years ago

12

- With the device starting from a still position (0 m/s), you drop it, allowing gravity to accelerate it towards the ground at 9

.8 m/s2. Assuming it takes 1.5 seconds for the device to hit the ground, what is it’s velocity? [V = a*t]

1 answer:

IrinaVladis [17]3 years ago

3 0

Starting velocity (u)=0
acceleration (a) = 9.8
time (t) = 1.5
final velocity (v) =?

v=at+u
v=9.8(1.5)+0
v=14.7m/s2

You might be interested in

What is buoyant force

Xelga [282]

Answer:

When an object is immersed in water. it is pulled downwards due to gravitational pull of earth. Water exerts upward force on the object. This makes object rise up. This upward force is called buoyancy or upthrust.

7 0

2 years ago

What would the acceleration of a car be from a stoplight if it started at 0 m/s and reached a

trapecia [35]

Answer:

The acceleration of a car would be: $a=12.8$ m/s²

Explanation:

Given

Initial velocity = $v_1=0$ m/s

Final velocity = $v_f=100$ m/s

Time elapsed = $t = 7.8$ s

To determine

We need to determine the acceleration of a car.

We know that acceleration is basically the rate of change in velocity over time.

Thus,

We can determine the acceleration using the formula

$\:\:\:a=\:\frac{v_f-v_i}{t}$

where

$a$ is the acceleration

$v_1$ is the initial velocity

$v_f$ is the final velocity

$t$ is time elapsed

now substituting the values $v_1=0$ , $v_f=100$ , and $t = 7.8$ in the formula

$\:\:\:a=\:\frac{v_f-v_i}{t}$

$a=\:\frac{100-0}{7.8}$

$a=\frac{100}{7.8}$

$a=12.8$ m/s²

Therefore, the acceleration of a car would be: $a=12.8$ m/s²

3 0

3 years ago

A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta

larisa [96]

Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

Horizontal displacement in 1.03 s

$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

6 0

3 years ago

Carlos is making phosphorous trichloride using the equation below. He adds 15 g of phosphorus.

Degger [83]

Given:

The balanced chemical reaction of the synthesis of phosphorus trichloride:

2P + 3Cl2 ===> 2PCl3

Initial amount of phosphorus = 15 grams

The amount of product produced from 15 grams of phosphorus:

15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3

The amount of chlorine is 44.31 grams, nearest to 45 grams.

8 0

4 years ago

Read 2 more answers

A plane flies 446 km east from city A to city B in 43.0 min and then 939 km south from city B to city C in 1.10 h. For the total

NeTakaya

Answer:

(a) Magnitude is 1039 km

(b) Direction of the displacement is $64.59^{\circ}$ South of East

(c) Average velocity magnitude is 570.88 km

(d) The direction of average velocity is $64.59^{\circ}$ South of East

(e) Average speed is 759.34 km/h

Solution:

Distance moved from A to B in East direction, $\vec{AB} = 446 km$

Distance moved from B to C in South direction, $\vec{BC} = - 939 km$

Time taken to move from A to B, t = 43.0 min = 0.72 h

Time taken to move from B to C, t' = 1.10 h

Now,

(a) The magnitude of displacement of the plane is provided by AC as shown in fig 1 and can be given as:

$AC = \sqrt{(AB)^{2} + (BC)^{2}}$

$AC = \sqrt{(446)^{2} + (- 939)^{2}}$ = 1039 km

(b) Direction of the displacement is given by:

$tan\theta = \frac{\vec{BC}}{\vec{AB}}$

$\theta = tan^{- 1}(\frac{- 939}{\vec{446}}) = - 64.59^{\circ}$

$64.59^{\circ}$ South of East

(c) Magnitude of the average speed is given by:

$v_{avg} = \frac{AC}{t + t'}$

$v_{avg} = \frac{1039}{1.82} = 570.88 km/h$

(d) The direction of the average velocity is the same as that of the displacement, i.e., $64.59^{\circ}$ South of East.

(e) The average speed of the [plane is given by:

$v'_{avg} = \frac{Total\ Distance\ Traveled}{Total\ Time}$

$v'_{avg} = \frac{446 + 939}{1.10 + 0.72} = 759.34 km/h$

6 0

3 years ago