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3 years ago

A 7-n vector at an angle of 45° to the horizontal has a vertical component that is about _______.

1 answer:

4 0

Vectors are easy once you break them down the right way.

Here is a shortcut formula i have got for you to use to find the x & y components of any vecor, once you know the magnitude and angle (theta):

to find the x component, use:
$[magnitude] * cos(theta)$

to find the y component use:
$[magnitude]*sin(theta)$

so for this question use:
$y = 7 * sin(45) = 4.94N$

You might be interested in

An object islaunched at velocity of 20 m/s in a direction making an angle of 25 degree upward with the horizontal. what is the m

Aneli [31]

Answer:

$\displaystyle y_m=3.65m$

Explanation:

Motion in The Plane

When an object is launched in free air with some angle respect to the horizontal, it describes a known parabolic path, comes to a maximum height and finally drops back to the ground level at a certain distance from the launching place.

The movement is split into two components: the horizontal component with constant speed and the vertical component with variable speed, modified by the acceleration of gravity. If we are given the values of $v_o$ and $\theta\\$ as the initial speed and angle, then we have

$\displaystyle v_x=v_o\ cos\theta$

$\displaystyle v_y=v_o\ sin\theta-gt$

$\displaystyle x=v_o\ cos\theta\ t$

$\displaystyle y=v_o\ sin\theta\ t -\frac{gt^2}{2}$

If we want to know the maximum height reached by the object, we find the value of t when $v_y$ becomes zero, because the object stops going up and starts going down

$\displaystyle v_y=o\Rightarrow v_o\ sin\theta =gt$

Solving for t

$\displaystyle t=\frac{v_o\ sin\theta }{g}$

Then we replace that value into y, to find the maximum height

$\displaystyle y_m=v_o\ sin\theta \ \frac{v_o\ sin\theta }{g}-\frac{g}{2}\left (\frac{v_o\ sin\theta }{g}\right )^2$

Operating and simplifying

$\displaystyle y_m=\frac{v_o^2\ sin^2\theta }{2g}$

We have

$\displaystyle v_o=20\ m/s,\ \theta=25^o$

The maximum height is

$\displaystyle y_m=\frac{(20)^2(sin25^o)^2}{2(9.8)}=\frac{71.44}{19.6}$

$\displaystyle y_m=3.65m$

7 0

2 years ago

A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc

iren2701 [21]

Answer:

The correct answer is;

The magnitude of the force is 35.12 N

Explanation:

To solve the question, we note that the friction is zero and the force causes motion of a stationary mass

One of the equations of motion is required such as

v² = u² + 2× a× s

Where

v = Final velocity = 5.93 m/s

u = Initial velocity = 0 m/s , object at rest

a = acceleration

s = distance moved = 32 meters

But v = Distance/Time = 32 m /5.4 s = 5.93 m/s

Therefore

5.93² = 2×a×32

or a = 35.12/ 64 = 0.55 m/s²

Therefore Force F = Mass m × Acceleration a

Where mass m = 64 kg

Therefore F = 64 kg×0.55 m/s² = 35.12 N

3 0

3 years ago

The horizontal surface on which the block slides is frictionless.The speed of the block before it touches the spring is 6.0 m/s.

Aloiza [94]

Answer:

V = 5.4 m/s

Explanation:

It is given that,

Let mass of the block, m = 10 kg

Spring constant of the spring, k = 2 kN/m = 2000 N/m

Speed of the block, v = 6 m/s

Compression in the spring, x = 15 cm = 0.15 m

Let V is the speed of the block moving at the instant the spring has been compressed 15 cm. It can be calculated using the conservation of energy of spring mass system.

$\dfrac{1}{2}m^2=\dfrac{1}{2}kx^2+\dfrac{1}{2}mV^2$

$mv^2=kx^2+mV^2$

$V^2=\dfrac{mv^2-kx^2}{m}$

$V^2=\dfrac{10\times 6^2-2000\times (0.15)^2}{10}$

V = 5.61 m/s

From the given options,

V = 5.4 m/s

Hence, this is the required solution.

3 0

3 years ago

A 7kg bowling ball is at rest at the end of a bowling lane. You push the ball with a force of 22N which induces a -1N of frictio

Valentin [98]

Explanation:

You can find the net force by summing the force components:

∑F = 22 N − 1 N

∑F = 21 N

Or by using Newton's second law:

∑F = ma

∑F = (7 kg) (3 m/s²)

∑F = 21 N

7 0

3 years ago

What is the distance from the moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are e

Ugo [173]

Answer:

the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is E)3.45 × 10⁸ m

Explanation:

The force that the Earth exerts on a mass m is

F_e = (G M_e m) / R_e²

where

G is the universal gravitational constant
M_e is the mass of Earth
R_e is the radius of Earth

The force that the Moon exerts on a mass m is

F_m = (G M_m m) / R_m²

where

G is the universal gravitational constant
M_m is the mass of the Moon
R_m is the radius of the Moon

Therefore, the point where the gravitational pulls of Earth and Moon are equal is:

F_e = F_m

R_e + R_m = R = 3.84×10⁸ m

Thus,

(G M_e m) / R_e² = (G M_m m) / R_m²

M_e / R_e² = M_m / (R - R_e²)

(R - R_e²) / R_e² = M_m / M_e

(R - R_e) / R_e = (M_m / M_e)^1/2

R_e(R/R_e -1) / R_e = (M_m / M_e)^1/2

R/ R_e = (M_m / M_e)^1/2 + 1

R_e = R / [(M_m / M_e)^1/2 + 1]

R_e = (3.84×10⁸ m) / [(7.35 x 10²² kg / 5.97 x 10²⁴ kg )^1/2 + 1]

R_e = 3.45 × 10⁸ m

Therefore, the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal is 3.45 × 10⁸ m.

4 0

3 years ago