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

A water skier is pulled behind a boat. When the skis push down on the water, what is the reaction force?

IrinaVladis [17]

The answer is C) The water pushed up on the skis

The water reacts to the downward force of the skis by pushing back up against the skis.

8 0

3 years ago

Read 2 more answers

Waxing or waning moon?

Dennis_Churaev [7]

Answer:

waning gibbious

Explanation:

7 0

2 years ago

An car is brought to rest in a distance of 652 m using a constant acceleration of -2.7 m/s2. What was the

Goshia [24]

Answer:

I’m so sorry I tried solving it but I don’t understand it can you explain the question a little bit more ty

Explanation:

7 0

3 years ago

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca

Aleks04 [339]

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$ . As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.

6 0

1 year ago

What safety feature melts to protect a circuit?

N76 [4]

A fuse melts to protect a circuit.

7 0

3 years ago