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3 days ago

5

A box weighing 46 newtons rests on an incline that makes an angle of 25° with the horizontal. What is the magnitude of the compo

nent of the box’s weight perpendicular to the incline?
(1)19 N
(2)21 N
(3)42 N
(4)46 N

1 answer:

inna [987]3 days ago

8 0

The force of the box’s weight acting perpendicular to the slope can be computed using the formula:

F = wcos(a)

In this equation, F represents the component of the weight perpendicular to the slope,

W denotes the box's total weight,

and A is the angle of the slope.

Thus, substituting values gives: F = (46)cos(25)

Resulting in F = 42 N

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To avoid an accident, a driver steps on the brakes to stop a 1000-kg car traveling at 65km/h. if the braking distance is 35 m, h

Maru [1056]

Initially, we need to calculate the acceleration required for the car to halt from its initial speed based on the distance traveled. This can be done using the formula,
2ad = Vf² - Vi²
where a represents acceleration, d is distance, and Vf and Vi are the final and initial speeds respectively. Plugging in the known quantities,
2(a)(35 m) = (0 m/s)² - ((65 km/h) x (1000 m/ 1 km) x (1 hr / 3600 s))²
The resulting acceleration is −4.66 m/s².
To calculate the force required to stop the car, we multiply the mass by the acceleration. This calculation yields -4,660 N, and we take the absolute value, which is 4,660 N.

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7 days ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

serg [1198]

Response:

a) The mug makes contact with the ground 0.7425m from the bar's end. b) |V|=5.08m/s θ= -72.82°

Clarification:

To address this issue, we begin with a diagram depicting the situation. (refer to the attached illustration).

a)

The illustration shows that the problem involves motion in two dimensions. To determine how far from the bar the mug lands, we need to find the time the mug remains airborne by examining its vertical motion.

To compute the time, we utilize the following formula with the known values:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

We have $y_{f}=0$ and $v_{y0}=0$ , allowing us to simplify the equation to:

$0=y_{0}+\frac{1}{2}at^{2}$

Now, we can calculate for t:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

We know $y_{0}=1.20m$ and $a=g=-9.8m/s^{2}$

The negative gravity indicates the downward motion of the mug. Hence, we substitute these values into the provided formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

Which results in:

t=0.495s

This time helps us evaluate the horizontal distance the mug traverses. Since:

$V_{x}=\frac{x}{t}$

Solving for x, we have:

$x=V_{x}t$

Substituting the known values yields:

x=(1.5m/s)(0.495s)

This calculates to:

x=0.7425m

b) With the time determining when the mug strikes the ground established, we can find the final velocity in the vertical direction using the formula:

$a=\frac{v_{f}-v_{0}}{t}$

The initial vertical velocity being zero simplifies our calculations:

$a=\frac{v_{f}}{t}$

Thus, we can determine the final velocity:

$V_{yf}=at$

Given that the acceleration equates to gravity (showing a downward effect), we substitute that alongside the previously found time:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

This leads to:

$V_{yf}=-4.851m/s$

Now, we ascertain the velocity components:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

Next, we find the speed by calculating the vector's magnitude:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

<pThus:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

Yielding:

|V|=5.08m/s

Lastly, to ascertain the impact direction, we apply the equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

<pFulfilling this provides:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

<pLeading to:

$\theta = -72.82^{o}$

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5 days ago

If the position of an object is zero at one instant, what is true about the velocity of that object?

ValentinkaMS [1149]

If the position of an object is zero at a particular moment, this does not provide any indication about its velocity. It might simply be moving through that point, and you observed it exactly when it was at zero.

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8 days ago

A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

Keith_Richards [1034]

To determine the average net force, we can calculate acceleration using:

x = 0.5*a*t^2

v = a*t

where x=3.6m and v=185 m/s.

Thus,

t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a

which gives us a= (0.5*v^2)/x

Since we have the known values of v and x, we can compute a by substituting these numbers.

The average net force is then given as:

F = m*a,

with m=7.5kg.

5 0

1 day ago

It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

Maru [1056]

Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

$F_X =64.65 \ N$

Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

Again substituting 42° for θ and 87.0° for $F$

$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

N = the normal force

Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

Therefore, the coefficient of friction is = 0.0984

5 0

6 days ago