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3 years ago

If a car travels at 45.0 m/s for 2 minutes. What is the car's distance?

Physics

2 answers:

NemiM [27]3 years ago

4 0

Converting two minutes into seconds gives us 120seconds

Distance=Average Speed*time
Distance=45/2*120
Distance =2700m

Rus_ich [418]3 years ago

3 0

Answer:

the answer is 90 m/s

Explanation:

.....

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An object has a mass of 10grams and a volume of 20m3. what is the density of the object

Pavlova-9 [17]

Explanation:

density = mass/ volume

in the question the mass is given in grams so will convert it into kg,

10 g = 0.01 kg

density= 0.01/ 20 = 1 / 2000 = 0.005 kg/m^3

7 0

3 years ago

Why is the reactor coolant water kept contained within the primary loop instead of allowing it to mix with the feed water and le

uysha [10]

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3 years ago

Read 2 more answers

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

$20 = 1.5125 v^2$

$v = 3.64 m/s$

Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

8 0

3 years ago

Freeeeeeeee poinnttttsssss

RUDIKE [14]

Answer:

Explanation:

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5 0

3 years ago

A dumbell has a mass of 95 kg. What force must be applied to accelerate it upward at 2.2 m/s2?

Sveta_85 [38]

A :-) F = ma
Given - m = 95 kg
a = 2.2 m/s^2
Solution -
F = ma
F = 95 x 2.2
F = 209

.:. The force is 209 N

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3 years ago