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2 years ago

What impulse acts on a car that is slowly coming to a stop?

Physics

2 answers:

iragen [17]2 years ago

7 0

Answer:

When cars bounce off each other, or rebound, there is a larger change in momentum and therefore a larger impulse

Delicious77 [7]2 years ago

6 0

Answer:

Explanation:

When cars bounce off each other, or rebound, there is a larger change in momentum and therefore a larger impulse. A larger impulse means that a greater force is experienced by the occupants of the cars. When cars crumple together, there is a smaller change in momentum and therefore a smaller impulse.

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A certain sea cow can paddle 2.0 m/s in still water. If she attempts to cross a river, from the south bank to the north with a c

aliina [53]

Answer:

v = 3.6m / s , θ = 56º

Explanation:

This is a relative speed exercise, let's use the Pythagorean theorem

v = √ (v₁² + v₂²)

where v₁ is the speed of the sea still water and v₂ the speed of the current

let's calculate

v = √ (2² + 3²)

v = 3.6m / s

to find the direction we use trigonometry

tan θ = v₂ / v₁

θ = tan⁻¹ (v₂ / v₁)

let's calculate

θ = tan⁻¹ (3/2)

θ = 56º

4 0

3 years ago

What will happen to the 0.1 N force if one charges is increased by a factor of 3?

AleksandrR [38]

Answer:

Explanation:

Force between two charges is given by the following expression

F = $\frac{KQ_1Q_2}{d^2}$ Q₁ and Q₂ are two charges and d is distance between two.

.1 = $\frac{KQ_1Q_2}{d^2}$

If Q₁ becomes three times , force will become 3 times . Hence force becomes .3 N in the first case.

Force F = .3 N

If charge becomes one fourth , force also becomes one fourth .

F= $\frac{.1}{4}$

= .025 N.

5 0

3 years ago

15. Sandra decided to talk a walk at her neighborhood park. She walks 20 meters

FinnZ [79.3K]

Answer:

Long question good luck:)..............

Explanation:

8 0

3 years ago

An object 2.7 cmtall is placed12 cmin front of a mirror. •What type of mirror and what radius of curvature are needed to createa

nasty-shy [4]

Answer:

a. Concave mirror, radius of curvature = 16 cm b. magnification = 2

Explanation:

a. Since the image is upright and larger than the object, we need a concave mirror.

We know image height, h'/object height, h = -image distance, d'/object distance, d

h'/h = -d'/d

Using the real is positive convention,

h'= + 5.4 cm, h = + 2.7 cm and d = + 12 cm.

So, + 5.4 cm/+ 2.7 cm = -d'/+ 12 cm

2 = -d'/12

d' = -2 × 12 cm

= -24 cm

Using the mirror formula 1/d + 1/d' = 2/r where r = radius of curvature of the mirror

1/+12 + 1/- 24 = 2/r

1/12(1 - 1/2) = 2/r

1/12(1/2) = 2/r

1/24 = 2/r

r/2 = 24

r = 2 × 24

r = 48 cm

magnification = image height, h'/object height,h = + 5.4 cm/+ 2.7 cm = 2

3 0

3 years ago

A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking

Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

$m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}$

The total motion has a total velocity and is

$Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}$

The time the worker take walking is

$t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s$

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

$x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m$

5 0

3 years ago