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2 years ago

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Physics

1 answer:

Anna [14]2 years ago

3 0

Answer:

Here, m=10 kg

The resultant force acting on the body is

F=(98N)2+(6N)2=10N

Let the resultant force F makes an angle θ w.r.t. 8N force.

From figure, tanθ=8N6N=43

The resultant acceleration of the body is

a=mF=10kg10N=1ms−2

The resulatnt acceleration is along the direction of the resulatnt force.

Hence, the resultant acceleration of the body is 1 ms−2 at an angle of tan−1(43) w.r.t. 8N force.

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There is a(n) __________ relationship between an object's weight and its mass.

Mkey [24]

There is a direct relationship between between an object's weight and its mass.

Weight is actually the force of gravity on the object's mass and it can be calculated by multiplying the object's mass and the acceleration due to gravity. The higher the mass, the higher the weight and the lower the mass, the lower the object's weight. Thus, the relationship is direct.

4 0

3 years ago

A jet is travelling at a speed of 1200 km/h and drops cargo from a height of 2.5 km above the ground Calculate the time it takes

OLEGan [10]

a) Time of flight: 22.6 s

To calculate the time it takes for the cargo to reach the ground, we just consider the vertical motion of the cargo.

The vertical position at time t is given by

$y(t) = h +u_y t - \frac{1}{2}gt^2$

where

h = 2.5 km = 2500 m is the initial height

$u_y = 0$ is the initial vertical velocity of the cargo

g = 9.8 m/s^2 is the acceleration of gravity

The cargo reaches the ground when

$y(t) = 0$

So substituting it into the equation and solving for t, we find the time of flight of the cargo:

$0 = h - \frac{1}{2}gt^2\\t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(2500)}{9.8}}=22.6 s$

b) 7.5 km

The range travelled by the cargo can be calculated by considering its horizontal motion only. In fact, the horizontal motion is a uniform motion, with constant velocity equal to the initial velocity of the jet:

$v_x = 1200 km/h \cdot \frac{1000 m/km}{3600 s/h}=333.3 m/s$

So the horizontal distance travelled is

$d=v_x t$

And if we substitute the time of flight,

t = 22.6 s

We find the range of the cargo:

$d=(333.3)(22.6)=7533 m = 7.5 km$

7 0

3 years ago

A diet is to contain at least 2400 mg vitamin C, 1800mg Calcium, and 1200 calories every day. Two foods, a dairy-based meal and

Kay [80]

Answer:

The answer is below

Explanation:

Let x represent the number of ounce of dairy based meal and let y represent the number of vegan option in ounce.

Since the diet must contain at least 2400 mg vitamin C, therefore:

50x + 20y ≥ 2400

Since the diet must contain at least 1800 mg Calcium, therefore:

30x + 20y ≥ 1200

Since the diet must contain at least 1200 calories, therefore:

10x + 40y ≥ 1200

Therefore the constraints are:

50x + 20y ≥ 2400

30x + 20y ≥ 1200

10x + 40y ≥ 1200

x > 0, y > 0

The graph was drawn using geogebra online graphing tool, and the solution to the problem is at:

C(30, 45) and D(48, 18)

dairy-based meal costs $0.042 per ounce and the vegan option costs $0.208 per ounce. The cost equation is:

Cost = 0.042x + 0.208y

At C(30, 45); Cost = 0.042(30) + 0.208(45) = $10.62

At C(48, 18); Cost = 0.042(48) + 0.208(18) = $5.76

The minimum cost is at (48, 18). That is 48 dairy based meal and 18 vegan

4 0

3 years ago

Which property do gas particles at the same temperature share?

atroni [7]

It’s potential energy

8 0

3 years ago

Read 2 more answers

Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co

aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

$r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m$

The radius is 1.195 m

Time period would be given by

$T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s$

Time period of the motion is 2.8375 s

Angular speed is given by

$\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s$

The angular speed of the motion is 2.21433 rad/s

4 0

3 years ago