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tia_tia [17]
3 years ago
13

An object has an acceleration of 6.0 m/s/s. if the net force acting upon this object were doubled, then its new acceleration wou

ld be _____ m/s/s.
Physics
1 answer:
Flauer [41]3 years ago
5 0
F has direct relation with a
then doubling F cause acc. to get double i:e 6×2=12
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To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r
sweet [91]

Answer:

a). 53.75 N and 101.92 N

b). 381.44 N and 723.25 N

Explanation:

V= 77 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 21.38 \frac{m}{s} \\V=106 \frac{km}{h}* \frac{1h}{3600 s} *\frac{1000m}{1 km} = 29.44 \frac{m}{s}

a).

ρ= 1.2 \frac{kg}{m^{3} }, A_{t}= 0.7 m^{2}, D_{t}= 0.28

F_{t1} = \frac{1}{2} * D_{t} * A_{t}* p_{t}* v_{t}^{2}

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 53.75 N

F_{t1} = \frac{1}{2} * 0.28 * 0.7m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 101.92 N

b).

ρ= 1.2 \frac{kg}{m^{3} }, A_{h}= 2.44 m^{2}, D_{h}= 0.57

F_{t1} = \frac{1}{2} * D_{h} * A_{h}* p_{h}* v_{h}^{2}

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 21.38^{2}= 381.44 N

F_{t1} = \frac{1}{2} * 0.57 * 2.44 m^{2} * 1.2\frac{kg}{m^{3} }* 29.44^{2}= 723.25 N

6 0
3 years ago
Which resultant force is NOT possible if 50 N force and a 60 N force act concurrently?
Sergio [31]

Answer:

The options are not shown, so i will answer in a general way.

Suppose the case where the forces act in opposite directions, then we need to subtract the forces, and we know that the magnitude of the resultant force will be:

60N - 50N = 10N

Now, suppose the case where both forces act in the exact same direction, in that case, we will add the forces to get:

60N + 50N = 110N

Then the only range of forces that we can get in this system, are the forces such:

10N ≤ F ≤ 110N

Any resultant force outside that range is not possible in this situation.

3 0
3 years ago
A sensational 130 kg midfielder has 695 J of
antoniya [11.8K]

Answer:

v = 3.27 m/s

Explanation:

KE = 1/2 mv^2

695 J = 1/2 (130kg)(v^2)

695 J / (1/2 x 130kg) = v^2

v^2 = square root of 10.69

v = 3.27 m/s

4 0
3 years ago
calculate the load placed 10m from the fulcrum that can be balanced by an effort of 5 N applied at a distance of 4 m from the fu
Elena L [17]

Answer:

A = 2 m from fulcrum

Explanation:

Product of anti clockwise  = Product of  clockwise moment

5 × 4 = 10 × A

20 = 10 x A

A = 20 / 10

A = 2 m from fulcrum

8 0
3 years ago
When your body is warmed by an electric blanket during the winter, this process is said to be
lana66690 [7]
The correct answer would be A. endothermic because endothermic is-<span>(of a reaction or process) accompanied by or requiring the absorption of heat. Hope this helps!:) If you need any more just tag me.

</span>
8 0
3 years ago
Read 2 more answers
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