Ask question

Ask question

All categories

3 years ago

6

At a certain location, a gravitational force with a magnitude of 350 newtons acts on a 70.-kilogram astronaut. What is the magni

tude of the gravitational field strength at this location? (1) 0.20 kg/N (2) 9.8 m/s2 (3) 5.0 N/kg (4) 25 000 Nkg

1 answer:

creativ13 [48]3 years ago

6 0

Answer:

3. 5.0N/kg

Explanation:

Gravitational field strength = gravitational force/mass of astronaut = 350N/70kg = 5.0N/kg

You might be interested in

PLZ HELP ME WITH THIS QUESTION! NO LINKS!!<br>will give 5 stars and brainliest

ohaa [14]

The answer is: Option 4

3 0

2 years ago

Read 2 more answers

FCFrdfAQefgEwdfgedfsaewfdedasfewreagargferdfares

Ivan

Answer: AAAAAAAAGGGGGHHHHJJJGSSSUUUUUUUUYCCFVGBHNJM

Explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET

6 0

3 years ago

Read 2 more answers

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A cons

Virty [35]

Complete Question

A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?

Answer:

$H_m=1.65m$

$H_E=1.16307m$

Explanation:

From the question we are told that

Mass of ball $M=2kg$

Length of string $L= 2m$

Wind force $F=13.2N$

Generally the equation for $\angle \theta$ is mathematically given as

$tan\theta=\frac{F}{mg}$

$\theta=tan^-^1\frac{F}{mg}$

$\theta=tan^-^1\frac{13.2}{2*2}$

$\theta=73.14\textdegree$

Max angle = $2*\theta= 2*73.14=>146.28\textdegree$

Generally the equation for max Height $H_m$ is mathematically given as

$H_m=L(1-cos146.28)$

$H_m=0.9(1+0.8318)$

$H_m=1.65m$

Generally the equation for Equilibrium Height $H_E$ is mathematically given as

$H_E=L(1-cos73.14)$

$H_E=0.9(1+0.2923)$

$H_E=1.16307m$

8 0

2 years ago

1. Which system has a higher random, average kinetic energy of its atoms, a glass of iced tea or a cup of hot tea?

natita [175]

The average kinetic energy of the particles is higher in the hot tea, so it also has greater thermal energy then the cold tea.

5 0

3 years ago

Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of 1.3 m/s parallel to the gro

pashok25 [27]

Answer:

Explanation:

cSep 20, 2010

well, since player b is obviously inadequate at athletics, it shows that player b is a woman, and because of this, she would not be able to hit the ball. The magnitude of the initial velocity would therefore be zero.

Anonymous

Sep 20, 2010

First you need to solve for time by using

d=(1/2)(a)(t^2)+(vi)t

1m=(1/2)(9.8)t^2 vertical initial velocity is 0m/s

t=.45 sec

Then you find the horizontal distance traveled by using

v=d/t

1.3m/s=d/.54sec

d=.585m

Then you need to find the time of player B by using

d=(1/2)(a)(t^2)+(vi)t

1.8m=(1/2)(9.8)(t^2) vertical initial velocity is 0

t=.61 sec

Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

v=.585m/.61 sec

so v=.959m/s

5 0

3 years ago

Read 2 more answers