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loris [4]
2 years ago
7

Can someone help me here. If the speed of an object does NOT change, the object is traveling at __________ speed.

Physics
2 answers:
Bond [772]2 years ago
7 0

Answer:

Constant Speed

Explanation:

Hope you pass have a nice day

RUDIKE [14]2 years ago
6 0
The answer is constant speed
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5 0
2 years ago
You work at a garden store for the summer, and you lift a 14 kg bag of fertilizer with a force of 227 N.
nika2105 [10]

Answer:

(a) Acceleration of the bag will be a=16.214m/sec^2  

(B) Weight of the bag will be 137.2 N

Explanation:

We have given mass of the bag m = 14 kg

Force with which bag is lifted = 227 N

(A) According to newtons law we force is equal to F = ma , here m is mass and a is acceleration

So 227=14\times a

a=16.214m/sec^2

(b) Acceleration due to gravity g=9.8m/sec^2

We know that weight is given by W = mg , here m is mass and g is acceleration due to gravity

So weight W=mg=14\times 9.8=137.2N

So weight of the bag will be 137.2 N

8 0
3 years ago
A 2.5 kg ball rolls forward at 10.0 m/s. What is the ball's momentum?​
Anuta_ua [19.1K]

Answer:25kgm/s

Explanation:

mass=2.5kg

Velocity=10m/s

Momentum=mass x velocity

Momentum=2.5 x 10

Momentum=25kgm/s

4 0
3 years ago
5. If a spring with a 40 N/m spring constant is used to push a 10 kg ball to a speed of 2 m/s,
kolbaska11 [484]

By compressing the spring a distance <em>x</em> (in m), you are storing 1/2 <em>k</em> <em>x</em> ² (in J) of potential energy, which is converted completely into kinetic energy 1/2 <em>m v</em> ², where

• <em>k</em> = 40 N/m = spring constant

• <em>m</em> = 10 kg = mass of the ball

• <em>v</em> = 2 m/s = ball's speed (at the moment the spring returns to its equilibrium point)

So we have

1/2 <em>k</em> <em>x</em> ² = 1/2 <em>m</em> <em>v</em> ²

<em>x</em> = √(<em>m</em>/<em>k</em> <em>v</em> ²) = √((10 kg) / (40 N/m) (2 m/s)²) = 1 m

3 0
2 years ago
A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
zhuklara [117]

The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

80 = 7θo

θo = 11.4°

Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

4 0
2 years ago
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