9. In the graph below, what is the force being exerted onthe 16-kg cart?
A. 4N
C. 16N
B. 8N
D. 32 N
9. In the graph below, what is the force being - 1

Answers

Answer 1
Answer:

Final answer:

The force being exerted on the 16-kg cart is 32N.

Explanation:

The force being exerted on the 16-kg cart can be determined using Newton's second law of motion, which states that force is equal to mass multiplied by acceleration. In the given graph, the cart is accelerating at a rate of 2 m/s2. Therefore, the force can be calculated as:

force = mass x acceleration

force = 16 kg x 2 m/s2 = 32 N

The 16-kg cart experiences a force of 32N, determined by applying Newton's second law of motion. According to this law, force equals mass multiplied by acceleration. In this scenario, the cart accelerates at a rate of 2 m/s². Substituting the values into the equation, the force exerted on the cart can be calculated as 16 kg multiplied by 2 m/s², resulting in a force of 32 N. This fundamental principle in physics establishes a quantitative relationship between force, mass, and acceleration.

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Answer 2
Answer:

Answer:

D

Explanation:


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A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the force the seat exerts against his body. Compare this with his weight using a ratio. (b) Calculate the direction and magnitude of the total force the seat exerts against his body.

Answers

Explanation:

It is given that,

Mass of the passenger, m = 75 kg

Acceleration of the rocket, a=49\ m/s^2

(a) The horizontal component of the force the seat exerts against his body is given by using Newton's second law of motion as :

F = m a

F=75\ kg* 49\ m/s^2

F = 3675 N

Ratio, R=(F)/(W)

R=(3675)/(75* 9.8)=5

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

F'=√(F^2+W^2)

F'=√((3675)^2+(75* 9.8)^2)

F' = 3747.7 N

The direction of force is calculated as :

\theta=tan^(-1)((W)/(F))

\theta=tan^(-1)((1)/(5))

\theta=11.3^(\circ)

Hence, this is the required solution.

Final answer:

The horizontal component of the force the seat exerts against the passenger's body is 3675 N. The ratio of this force to the passenger's weight is 5. The total force the seat exerts has a magnitude of 3793 N.

Explanation:

(a) To calculate the horizontal component of the force the seat exerts against the passenger's body, we can use Newton's second law, which states that force is equal to mass times acceleration. In this case, the mass of the passenger is 75.0 kg and the acceleration of the rocket sled is 49.0 m/s2. So the force exerted by the seat is:

Force = mass * acceleration

Force = 75.0 kg * 49.0 m/s2

Force = 3675 N

Now let's compare this force to the passenger's weight. The weight of an object is given by the formula:

Weight = mass * gravitational acceleration

Weight = 75.0 kg * 9.8 m/s2

Weight = 735 N

To find the ratio, we divide the force exerted by the seat by the weight of the passenger:

Ratio = Force / Weight

Ratio = 3675 N / 735 N

Ratio = 5

(b) The total force the seat exerts against the passenger's body has both a horizontal and vertical component. The direction of the total force is the same as the direction of the acceleration of the rocket sled. The magnitude of the total force can be found using the Pythagorean theorem:

Total Force = √(horizontal component2 + vertical component2)

Total Force = √(36752 + 7352)

Total Force = 3793 N

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The 160-lblb crate is supported by cables ABAB, ACAC, and ADAD. Determine the tension in these wire

Answers

Answer:

F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004

Explanation:

Using the diagram (see attachment) we extract the following position vectors:

Vector (OA) = 6i + 0j + 0k \nVector (OB) = 0i + 3j + 2k \nVector (OC) = 0i - 2j + 3k

Next step is to find unit vectors u_(AB) ,u_(AC), u_(AD), u_(AE) as follows:

u_(AB) = (vector(AB))/(magnitude(AB)) \n= \frac{OB - OA}{magnitude({vector(OB - OA))} }\n=(-6i +3j+2k)/(√(6^2 + 3^2+2^2) ) \n\n=-0.857 i +0.429j+0.286k\n\nu_(AC) = (vector(AC))/(magnitude(AC)) \n= \frac{OC - OA}{magnitude({vector(OC - OA))} }\n=(-6i -2j-3k)/(√(6^2 + 2^2+3^2) ) \n\n=-0.857 i -0.286j+0.429k\n\nu_(AD) = +1i\nu_(AC) = -1k

Using the diagram we find the corresponding vectors Forces:

F_(AB) = F_(AB) i + F_(AB)j +F_(AB)k\nF_(AC) = F_(AC) i + F_(AC)j +F_(AC)k\nF_(AD) = F_(AD) i + F_(AD)j +F_(AD)k\nW = -160 k

Equation of Equilibrium:

Sum of forces = 0\nF_(AB). u_(AB) + F_(AC).u_(AC) + F_(AD).u_(AD) + W = 0\n(-0.857F_(AB)i + 0.429F_(AB)j +0.286F_(AB)k) + (-0.857F_(AC)i - 0.286F_(AC)j +0.429F_(AC)k) + (+1F_(AD) i)  + (-160k) = 0

Comparing i , j and k components as follows:

-0.857F_(AB) -0.857F_(AC)  +1F_(AD)  = 0\n+ 0.429F_(AB) - 0.286F_(AC) = 0\n+0.286F_(AB) +0.429F_(AC)  =  160

Solving Above equation simultaneously we get:

F_(AB) = 172.1356\nF_(AC) = 258.2033\nF_(AD) = 368.8004

The aorta pumps blood away from the heart at about 40 cm/s and has a radius of about 1.0 cm. It then branches into many capillaries, each with a radius of about 5 x 10−4 cm carrying blood at a speed of 0.10 cm/s.How many capillaries are there?

Answers

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

A_1v_1=nA_2v_2               (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

A=\pi r^2\n\nA_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\n\nA_2=\pi r_2^2=\pi (5*10^(-4)cm)^2=7.85*10^(-7)cm^2

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

n=(A_1v_1)/(A_2v_2)=((3.1415cm^2)(40cm/s))/((7.85*10^(-7)cm^2)(0.10cm/s))=1.6*10^9

Then, the number of capillaries is 1.6*10^9

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

A spectroscope:measures light from distant objects
makes object far away look closer
receives radio signals from objects in space

Answers

Answer:

Option A

Measures light from distant objects

Explanation:

A spectroscope is used to measure the use of light from a distant object to work out the object is made of.

It could be the single-most powerful tool astronomers use.

Professor Fred Watson from the Australian Astronomical Observatory says that "It lets you see the chemicals being absorbed or emitted by the light source"

What is the force (in Newtons, 1 Newton = 1Kgm/s2) required to accelerate a 1500 Kg car to 3 m/s2?

Answers

Answer:

F=4500N

Explanation:

F=m×g

F=1500kg×3m/s²

F=4500N

Answer:

F=4500N

Explanation:

F=m×g

F=1500kg×3m/s²

F=4500N