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Luba_88 [7]
2 years ago
7

Which of the following objects exerts a gravitational force?

Physics
2 answers:
klio [65]2 years ago
7 0
Answer: the earth
hope this helps
kompoz [17]2 years ago
4 0

the earth exerts a gravitational force

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A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

Unbalanced Force = Horizontal Component Applied Force - Frictional Force

Unbalanced Force = Fx - f

But, from Newtons Second Law of Motion:

Unbalanced Force = ma

comparing the equations:

ma = Fx - f

f = F Cos θ - ma

where,

f = frictional force  = ?

F = Applied force  = 593 N

m = mass of person = 49 kg

a = acceleration = 0.57 m/s²

θ = Angle with horizontal = 30°

Therefore,

f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)

f = 513.55 N - 27.93 N

<u>f = 485.62 N</u>

6 0
3 years ago
if a bowling ball hits a wall a force of 6 N, the wall exerts a force of how much back. on the bowling ball
Ray Of Light [21]
There has to be 6 N for a reactant


5 0
3 years ago
An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 60 cm long and has a mass of 3.8 kg, with the center of
Serggg [28]

Answer:

(a) τ = 26.58 Nm

(b) τ = 18.79 Nm

Explanation:

(a)

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

d₁ = perpendicular distance between ball and shoulder = 60 cm = 0.6 m

τ₁ = (29.4 N)(0.6 m)

τ₁ = 17.64 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of 60 cm = (0.4)(60 cm) = 24 cm = 0.24 m

τ₂ = (37.24 N)(0.24 m)

τ₂ = 8.94 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 17.64 Nm + 8.94 Nm

<u>τ = 26.58 Nm</u>

<u></u>

(b)

Now, the arm is at 45° below horizontal line.

First we find the torque due to the ball in hand:

τ₁ = F₁d₁

where,

τ₁ = Torque due to ball in hand = ?

F₁ = Force due to ball in hand = m₁g = (3 kg)(9.8 m/s²) = 29.4 N

42.42 cm = 0.4242 m

τ₁ = (29.4 N)(0.4242 m)

τ₁ = 12.47 Nm

Now, we calculate the torque due to the his arm:

τ₁ = F₁d₁

where,

τ₂ = Torque due to arm = ?

F₂ = Force due to arm = m₂g = (3.8 kg)(9.8 m/s²) = 37.24 N

d₂ = perpendicular distance between center of mass and shoulder = 40% of (60 cm)(Cos 45°) = (0.4)(42.42 cm) = 16.96 cm = 0.1696 m

τ₂ = (37.24 N)(0.1696 m)

τ₂ = 6.32 Nm

Since, both torques have same direction. Therefore, total torque will be:

τ = τ₁ + τ₂

τ = 12.47 Nm + 6.32 Nm

<u>τ = 18.79 Nm</u>

3 0
2 years ago
A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe
Lina20 [59]

The final speed of the block after the collision with the obstacle is \boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The mass of the block is 6.0\,{\text{kg}}.

The initial momentum of the block is 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}.

The impulse imparted by the obstacle is 10\,{\text{N}} \cdot {\text{s}}.

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

{p_f} - p{ & _i} = I               …… (1)                                    

Substitute 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} for {p_i} and - 10\,{\text{N}} \cdot {\text{s}} for I  in equation (1).

 \begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

The final momentum of the block can be expressed as:

{p_f} = m{v_f}                   …… (2)                                  

Substitute 20\text{kg}\;\text{m/s} for {p_f} and 6.0\,{\text{kg}} for m in equation (2).

 \begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}

Thus, the final speed of the block after the collision with the obstacle is \boxed{3.33\;\text{m/s}}.

Learn More:

  1. Choose the 200 kg refrigerator. Set the applied force to 400 n (to the right) brainly.com/question/4033012
  2. With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward brainly.com/question/9719731
  3. Which of the following is an example of a nonpoint source of freshwater pollution brainly.com/question/1482712

Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords:  Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

5 0
3 years ago
Read 2 more answers
Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba
mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0
2 years ago
Read 2 more answers
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