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If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choi

vivado [14]

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

= 5 N

Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

Therefore, the information given is not enough to determine the acceleration of the box.

8 0

3 years ago

Two fish swimming in a river have the following equations of motion:

IRINA_888 [86]

Answer:

The second fish, X2, is moving faster than the first fish, X1

Explanation:

The given parameters for the equation of motion of the fishes are;

X1 = -6.4 m + (-1.2 m/s)×t

X2 = 1.3 m + (-2.7 m/s)×t

The given equation are straight line equations in the slope and intercept form, where the slope is the speed and in m/s and the intercept is the starting point of swimming of the fishes

For the first fish, the intercept = -6.4 m, the slope = the speed = -1.2 m/s

For the second fish, the intercept = 1.3 m, the slope = the speed = -2.7 m/s

Whereby the fishes are swimming in the opposite direction of the measurement of length, we have;

The magnitude of the speed of the second fish $\left | -2.7 \ m/s \right | = 2.7 \ m/s$ , is larger than the magnitude of the speed of the first fish $\left | -1.2 \ m/s \right | = 1.2 \ m/s$