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Elza [17]
3 years ago
7

A truck accelerates on a highway from rest to 216 m/s in 10s. What is the magnitude of acceleration?

Physics
1 answer:
Serhud [2]3 years ago
6 0

Explanation:

acceleration is the change in velocity per time taken

a = ∆v/t.

a = (216 - 0)/10

a = 216/10

a = 21.6m/s²

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An object weighs 10N on earth .what is the objects weight on a planet one tenth the earths mass and one half its radius?
earnstyle [38]
We know, weight = mass * gravity 
10 = m * 9.8
m = 10/9.8 = 1.02 Kg

Now, Let, the gravity of that planet = g'
g' = m/r²   [m,r = mass & radius of that planet ]
g' = M/10 / (1/2R)²   [M, R = mass & radius of Earth ]
g' = 4M / 10R²
g' = 2/5 * M/R²
g' = 2/5 * g   
g' = 2/5 * 9.8
g' = 3.92

Weight on that planet = planet's gravity * mass
W' = 3.92 * 1.02
W' = 4 N

In short, Your Answer would be 4 Newtons

Hope this helps!
3 0
3 years ago
A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
koban [17]

<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

<u>Explanation:</u>

  Let the east point towards positive X-axis and north point towards positive Y-axis.

  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

    Angle from positive horizontal axis = 86.85⁰

 So Velocity of the dog relative to the road = 26.04 m/s 86.85⁰ east of north = 26.04 m/s 3.15⁰ north of east.

4 0
3 years ago
GIVEAWAY IMAKE SURE YOU SAY TY! ILL GIVE MORE IF DO
Gwar [14]

Answer:

ok

Explanation:

ok

6 0
2 years ago
Read 2 more answers
Studen
Vesna [10]

Complete Question

A football coach walks 18 meters westward, then 12 meters eastward, then 28 meters westward, and finally 14 meters eastward.

a

From this motion what is the distance covered

b

What is the magnitude and direction of the displacement

Answer:

a

  D =   72 \  m

b

Magnitude

             d =  - 20 \ m

 Direction  

West

Explanation:

From the question we are told that

The first distance covered westward is d_w_1  =  18 \  m /tex]     The  first distance covered eastward is [tex]d_e1 =  12 \  m /tex]      The second distance covered westward is [tex]d_w_2  =  28 \  m /tex]      The  second distance covered eastward is [tex]d_e2 =  14 \  m /tex]   Generally the distance covered is mathematically represented as       [tex]D =  d_w1 + d_w2 + d_e1 + d_e2

=> D =   18 + 28 + 12 +  14

=> D =   72 \  m

For the second question eastward is in the direction of the positive x-axis so it would be positive and westward is in the direction of the negative x-axis so it would be negative

The magnitude of the displacement is

d =  -d_w1 -d_w2 + d_e1 + d_e2

=>  d =  -18-28 + 12 + 14

=> d =  - 20 \ m

The direction is west

7 0
3 years ago
Help with question 16 and 17
storchak [24]
<h3>16.</h3>

Your answer is correct.

___

<h3>17.</h3>

The fractional change in resistance is equal to the given temperature coefficient multiplied by the change in temperature.

  R = R₀×(1 + α×ΔT)

  R = (10.0 Ω)×(1 + 0.004×(65 -20)) = 11.8 Ω

5 0
3 years ago
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