How many significant figures? 5.0001 O None of these are correct O 5 02 0 1

NEED ANSWER IMMEDIANTELY What is the net force on this object? A) 0N it is balanced B) 200N down C) 400N up D) 600N down

yaroslaw [1]

The answer is B, when you subtract 400N from 600N, you are left with 200N in net force.

4 0

3 years ago

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At a baseball game, the batter hit a fly ball at time t = 0 s. The outfielder caught the ball at t = 5.8 s. When was the ball at

Agata [3.3K]

We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.

8 0

3 years ago

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Help asap please I will give you 5stars

boyakko [2]

Explanation:

In the parallel combination, the equivalent resistance is given by :

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+....$

4. When three 150 ohms resistors are connected in parallel, the equivalent is given by :

$\dfrac{1}{R}=\dfrac{1}{150}+\dfrac{1}{150}+\dfrac{1}{150}\\\\R=50\ \Omega$

5. Three resistors of 20 ohms, 40 ohms and 100 ohms are connected in parallel, So,

$\dfrac{1}{R}=\dfrac{1}{20}+\dfrac{1}{40}+\dfrac{1}{100}\\\\=11.76\ \Omega$

Hence, this is the required solution.

6 0

3 years ago

You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has

vovikov84 [41]

Answer:8.75 s,

136.89 m

Explanation:

Given

Initial velocity $=70 mph\approx 31.29 m/s$

velocity after 5 s is $30 mph\approx 13.41 m/s$

Therefore acceleration during these 5 s

$a=\frac{v-u}{t}$

$a=\frac{13.41-31.29}{5}=-3.576 m/s^2$

therefore time required to stop

v=u+at

here v=final velocity =0 m/s

initial velocity =31.29 m/s

$0=31.29-3.576\times t$

$t=\frac{31.29}{3.576}=8.75 s$

(b)total distance traveled before stoppage

$v^2-u^2=2as$

$0^2-31.29^2=2\times (-3.576)\cdot s$

s=136.89 m

3 0

2 years ago

A 1.5m long string weighs 0.0020 kg. It is tensioned to 100N. A disturbance travels along it with a wavelength of 1.5m, find:a)

Zigmanuir [339]

Answer:

the propagation velocity of the wave is 274.2 m/s

Explanation:

Given;

length of the string, L = 1.5 m

mass of the string, m = 0.002 kg

Tension of the string, T = 100 N

wavelength, λ = 1.5 m

The propagation velocity of the wave is calculated as;

$v = \sqrt{\frac{T}{\mu} } \\\\\mu \ is \ mass \ per \ unit \ length \ of \ the \ string\\\\\mu = \frac{0.002 \ kg}{1.5 \ m} = 0.00133 \ kg/m\\\\v = \sqrt{\frac{100}{0.00133} } \\\\v = 274.2 \ m/s$

Therefore, the propagation velocity of the wave is 274.2 m/s

7 0

2 years ago