With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why?

Calculate the magnitude of the total impulse applied to the car to bring it to rest

n200080 [17]

Answer:

Total impulse = $mv$ = Initial momentum of the car

Explanation:

Let the mass of the car be 'm' kg moving with a velocity 'v' m/s.

The final velocity of the car is 0 m/s as it is brought to rest.

Impulse is equal to the product of constant force applied to an object for a very small interval. Impulse is also calculated as the total change in the linear momentum of an object during the given time interval.

The magnitude of impulse is the absolute value of the change in momentum.

$|J|=|p_f-p_i|$

Momentum of an object is equal to the product of its mass and velocity.

So, the initial momentum of the car is given as:

$p_i=mv$

The final momentum of the car is given as:

$p_f=m(0)=0$

Therefore, the impulse is given as:

$|J|=|p_f-p_i|=|0-mv|=|-mv|=mv$

Hence, the magnitude of the impulse applied to the car to bring it to rest is equal to the initial momentum of the car.

5 0

3 years ago

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Varvara68 [4.7K]

Answer:

Explanation:

A force of $55\text{ }N$ is acting on a wagon along the road. The wagon weights $7.5\text{ }kg$ . Acceleration of the wagon is given as $5.3\text{ }\frac{m}{s^{2}}$ .

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon $\times$ Acceleration of wagon

$F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N$

$F_{friction}$ has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0

3 years ago

Um objeto de 4cm de altura está a 30cm de um espelho côncavo, cujo raio de curvatura tem valor absoluto de 20cm.

Shkiper50 [21]

a) The distance of the image from the mirror is 15 cm

b) The size of the image is -2 cm (inverted)

Explanation:

a)

We can solve this first part of the problem by applying the mirror equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

For a mirror, the focal length is half the radius of curvature, R:

$f=\frac{R}{2}$

For this mirror, R = 20 cm, so its focal length is

$f=\frac{20}{2}=+10 cm$ (positive for a concave mirror)

Here we also know:

p = 30 cm is the distance of the object from the mirror

So, by applying the equation, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{10}-\frac{1}{30}=\frac{1}{15} \rightarrow q = 15 cm$

b)

We can solve this part by using the magnification equation:

$M=-\frac{y'}{y}=\frac{q}{p}$

where

y' is the size of the image

y is the size of the object

q is the distance of the image from the mirror

p is the distance of the object from the mirror

Here we have:

q = 15 cm

p = 30 cm

y = 4 cm

Solving for y', we find the size of the image:

$y'=-y\frac{q}{p}=-(4)\frac{15}{30}=-2 cm$

and the negative sign means that the image is inverted.

#LearnwithBrainly

6 0

3 years ago

viva [34]

An LDR's resistance changes with light intensity, while a thermistor's resistancce changes with temperature.

In dark, LDR's resistance is large and in the day/light LDR's resistance is small.

At low temperature, thermistor's resistance is large, while at large temperature it resistance is small.

In an LDR Resistance increases as light intensity falls, while in a thermistor resistance falls as temperature falls.

5 0

2 years ago

Two crates, one with mass 5.4 kg and the other with mass 8.2 kg, connected by a light rope. The coefficient of kinetic friction

Gnom [1K]

Answer:

R= 2.5 :ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks

Explanation:

We apply Newton's second law:

∑F=m*a

velocity is constant ,then , a=0

Nomenclature

W: weight

m: mass

N : normal force

Ff: Friction force

μk: coefficient of kinetic friction

T: tension force in the rope

F: applied horizontal force

g: acceleration due to gravity.

Force Calculation

W₁=m₁*g=5.4 kg *9.8m/s²=52.92 N

W₂=m₂*g=8.2 kg *9.8m/s²= 80.36N

∑Fy=0

N₁-W₁=0 , N₁=W₁ = 52.92 N

N₂-W₂=0, N₂=W₂=80.36N

Ff₁= μk* N₁=0.4*52.92 N = 21.16N

Ff₂= μk* N₂=0.4*80.36N = 32.14N

Look at the attached graphic

Free-Body diagram m₁=5.4 kg

∑Fx=0

T- Ff₁=0 , T= Ff₁ , T= 21.16N

Free-Body diagram m₂=8.2 kg

∑Fx=0

F-T- Ff₂=0 , F=T+Ff₂= 21.16N+32.14N=53.3N

Ratio of the magnitude of the applied horizontal force to the magnitude of the tension in the rope connecting the blocks (R)

R= F/T= 53.3N/21.16N = 2.5

3 0

3 years ago