All categories

3 years ago

Can someone please help me because this is a hard problem to solve.

Physics

1 answer:

ivolga24 [154]3 years ago

7 0

Answer:

25.0 m / 10 m/s = 2.5 s

50.0 / 9.5 = 5.26 s

25.0 / 11.1 = 2.25 s

T (Wood) = 2.5 + 5.26 + 2.25 = 10.0 s

Mrs Wood runs 10 s vs 10.4 for Mr Overstreet

You might be interested in

Using newtons second law of motion, how fast for 100 KG object accelerates 350 N of force is applied to

fenix001 [56]

Answer:

3.5m/s^2

Explanation:

From Newton's second Law of Motion

F = ma

Where F is the applied force, m is the mass of the object and a is the acceleration.

F = 350 N

Mass = 100kg

350N = 100×a

a = 350/100

a = 3.5m/s^2

The acceleration of the object will be 3.5m/s^2

6 0

3 years ago

What does a physicist study?

MissTica

Answer:

A. Matter and energy

Explanation:

5 0

3 years ago

a chuck wagon with an initial velocity of 4 m/s and a mass of 35 kg gets a push with 350 joules of force. what is the wagon's fi

Kitty [74]

Answer:

the final velocity of the wagon is 6 m/s.

Explanation:

Given;

initial velocity of the wagon, u = 4 m/s

mass of the wagon, m = 35 kg

energy applied to the wagon, E = 350 J

The final velocity of the wagon is calculated as;

E = ¹/₂m(v² - u²)

$m(v^2-u^2) = 2E\\\\v^2-u^2 = \frac{2E}{m} \\\\v^2 = \frac{2E}{m} + u^2\\\\v = \sqrt{\frac{2E}{m} + u^2} \\\\v = \sqrt{\frac{2(350)}{35} + (4)^2}\\\\v = 6 \ m/s$

Therefore, the final velocity of the wagon is 6 m/s.

8 0

3 years ago

How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec

Rus_ich [418]

Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

$\text{ Initial vertical velocity = 0 m/s}$

let's find how long the ball remained in the air.

$\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}$

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

$\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}$

6 0

1 year ago

What type of galaxy consists of nebulas therefore has younger stars?

ss7ja [257]

Answer:

spiral galaxies

Hope this helps

8 0

3 years ago

Read 2 more answers