Ask question

Ask question

All categories

3 years ago

10

A nurse pushes a cart by exerting a force on the handle at a downward angle of 36 degrees below the horizontal. the loaded cart

has a mass of 28kg, and the force of friction is 65 n. what is the magnitude of force the nurse must exert to move at a constant velocity in newtons?

1 answer:

Troyanec [42]3 years ago

4 0

Since it was stated that it must move at constant velocity, so the only force it must overpower is the frictional force.

So the equation is:

F cos θ = Ff

F cos 36 = 65 N

F = 80.34 N

<span>So the nurse must exert 80.34 N of force</span>

You might be interested in

How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?

Andrew [12]

Answer:

Explanation:

There's a formula for this:

$F = k*displacement$

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

3 0

3 years ago

If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g

EastWind [94]

Answer:

Decreased by a factor of 4.5

Explanation:

"We have Newton formula for attraction force between 2 objects with mass and a distance between them:

$F_G = G\frac{M_1M_2}{R^2}$

where $G =6.67408 × 10^{-11} m^3/kgs^2$ is the gravitational constant on Earth. $M_1, M_2$ are the masses of the object and Earth itself. and R distance between, or the Earth radius.

So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:

$\frac{F_G}{f_g} = \frac{G\frac{M_1M_2}{R^2}}{G\frac{M_1m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{\frac{M_2}{R^2}}{\frac{m_2}{r^2}}$

$\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}$

$\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2$

Since $M_2 = 2m_2$ and $r = R/3$

$\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5$

So gravity would have been decreased by a factor of 4.5

8 0

3 years ago

A bullet fired horizontally over level ground hits the ground in 0.5 second. If it had been fired with twice the speed in the sa

AnnZ [28]

Answer:

In 0.5 seconds.

Explanation:

The time would be the same because it only depends on the height and the vertical component of the initial velocity. This is of course because each direction must be treated independently. Since between both cases only the horizontal speed changes, the height is the same and the vertical component of the initial velocity is null for both, the time to fall is the same.

8 0

2 years ago

An electron of mass 9.11×10−31 kgkg leaves one end of a TV picture tube with zero initial speed and travels in a straight line t

ki77a [65]

Explanation:

We will use the equations of constant acceleration to find out $a_{x}$ and time t.

As we know that the initial speed is zero. So

(a)

$v_{0x} = 0$

$x - x_{o} = 1.25$ × $10^{-2}$ m

$v_{x} = 3.3$ × $10^{6}$ m/s

$v^{2} _{x} = v^{2} _{x_{o} } + 2a_{x} (x - x_{o} )$

$a_{x} = \frac{v^{2} _{x} - v^{2} _{ox} }{2(x - x_{o}) }$

= $\frac{(3.3 * 10^{6})^{2} - 0 }{2(1.25 * 10^{-2}) }$

= 4.356× $10^{14}$ m/s²

(b)

$v_{x} = v_{ox} + a_{x}t$

$t = v_{x} - vo_{x}/a_{x}$

$t = \frac{3.00 * 10^{6} }{4.356*10^{14} }$ = 6.8870× $10^{-9}$ s

(c)

Σ $F_{x} = ma_{x}$

= (9.11× $10^{-31}$ )(4.356× $10^{14}$ m/s²)

= 3.968× $10^{-16}$ N

6 0

2 years ago

It takes 1 minute for 45 c to pass a point in a circuit, what is the current flowing through the circuit?

ohaa [14]

Answer:

0.75 A

Explanation:

An electric current is a flow of charged particles.

A current is defined through its intensity, which is given by:

$I=\frac{q}{t}$

where

I is the current intensity

q is the charge passing through a given point in the circuit

t is the time it takes for the charge q to pass that point in the circuit

In this problem, we have:

q = 45 C is the charge passing through the point in the circuit

$t=1 min = 60 s$ is the time elapsed

Therefore, the current intensity is:

$I=\frac{45}{60}=0.75 A$

5 0

3 years ago