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2 years ago

How are the Earth's crust and jigsaw alike?

2 answers:

4 0

The earth's surface is composed tectonic plates of basaltic oceanic crust with lighter granitic continental crust floating in isostatic balance on the oceanic crust.<span>
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Blizzard [7]2 years ago

4 0

The earths surface is composed tectonic plates of basaltic oceanic crust with lighter granitic continental crust floating in isostatic balance on the oceanic crust.

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50 N of force is exerted on a crate. It accelerates at a rate of 1.2 m/s/s. Calculate its mass

vitfil [10]

Answer:

m = F/a = 50 / 1.2 = 41.6666.... 42 kg

Explanation:

4 0

2 years ago

Give three practical uses of electromagnets

PtichkaEL [24]

Three practical uses of electromagnets would be
Light. The waves
Electric locks. The bolts that are used by the
coils
Cranes. The electric ones that are found in
scraps
Cars computers washing machines all use
Electromagnets . Even CD players DVDs
Players

7 0

3 years ago

A car initially going 61 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 7 seconds.

Bond [772]

Answer:

See the attachment below for the graphics in part (a)

The initial velocity for this time interval is u = 61ft/sec and the final velocity is 0m/s because the car comes to a stop.

This a constant acceleration motion considering the given time interview over which the brakes are applied. So the equals for constant acceleration motion apply here.

Explanation:

The full solution can be found in the attachment below.

Thank you for reading. I hope this post is helpful to you.

4 0

2 years ago

A 40.0 kg wheel, essentially a thin hoop with radius 0.810 m, is rotating at 438 rev/min. It must be brought to a stop in 21.0 s

garik1379 [7]

Answer:

(a) Workdone = -27601.9J

(b) Average required power = 1314.4W

Explanation:

Mass of hoop,m =40kg

Radius of hoop, r=0.810m

Initial angular velocity Winitial=438rev/min

Wfinal=0

t= 21.0s

Rotation inertia of the hoop around its central axis I= mr²

I= 40 ×0.810²

I=26.24kg.m²

The change in kinetic energy =K. E final - K. E initail

Change in K. E =1/2I(Wfinal² -Winitial²)

Change in K. E = 1/2 ×26.24[0-(438×2π/60)²]

Change in K. E= -27601.9J

(a) Change in Kinetic energy = Workdone

W= 27601.9J( since work is done on hook)

(b) average required power = W/t

=27601.9/21 =1314.4W

4 0

2 years ago

Read 2 more answers

The maximum distance from the Earth to the Sun (at aphelion) is 1.521 1011 m, and the distance of closest approach (at perihelio

LUCKY_DIMON [66]

Answer:

29274.93096 m/s

$2.73966\times 10^{33}\ J$

$-5.39323\times 10^{33}\ J$

$2.56249\times 10^{33}\ J$

$-5.21594\times 10^{33}$

Explanation:

$r_p$ = Distance at perihelion = $1.471\times 10^{11}\ m$

$r_a$ = Distance at aphelion = $1.521\times 10^{11}\ m$

$v_p$ = Velocity at perihelion = $3.027\times 10^{4}\ m/s$

$v_a$ = Velocity at aphelion

m = Mass of the Earth = 5.98 × 10²⁴ kg

M = Mass of Sun = $1.9889\times 10^{30}\ kg$

Here, the angular momentum is conserved

$L_p=L_a\\\Rightarrow r_pv_p=r_av_a\\\Rightarrow v_a=\frac{r_pv_p}{r_a}\\\Rightarrow v_a=\frac{1.471\times 10^{11}\times 3.027\times 10^{4}}{1.521\times 10^{11}}\\\Rightarrow v_a=29274.93096\ m/s$

Earth's orbital speed at aphelion is 29274.93096 m/s

Kinetic energy is given by

$K=\frac{1}{2}mv_p^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(3.027\times 10^{4})^2\\\Rightarrow K=2.73966\times 10^{33}\ J$

Kinetic energy at perihelion is $2.73966\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_p}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.471\times 10^{11}}\\\Rightarrow P=-5.39323\times 10^{33}$

Potential energy at perihelion is $-5.39323\times 10^{33}\ J$

$K=\frac{1}{2}mv_a^2\\\Rightarrow K=\frac{1}{2}\times 5.98\times 10^{24}(29274.93096)^2\\\Rightarrow K=2.56249\times 10^{33}\ J$

Kinetic energy at aphelion is $2.56249\times 10^{33}\ J$

Potential energy is given by

$P=-\frac{GMm}{r_a}\\\Rightarrow P=-\frac{6.67\times 10^{-11}\times 1.989\times 10^{30}\times 5.98\times 10^{24}}{1.521\times 10^{11}}\\\Rightarrow P=-5.21594\times 10^{33}$

Potential energy at aphelion is $-5.21594\times 10^{33}\ J$

6 0

2 years ago